💡Okay, so I have this equation here: ( x^4 - 4x^3 + 6x^2 - 40x - 64 = 0 ). The problem is asking me to find the pure imaginary solutions. Hmm, pure imaginary solutions mean that the solutions are of the form ( x = ki ), where ( k ) is a real number and ( i ) is the imaginary unit. Alright, so I remember that if ( x ) is purely imaginary, then substituting ( x = ki ) into the equation should satisfy it. Let me try that. So, I'll substitute ( x = ki ) into each term of the equation.First term: ( (ki)^4 ). Let me compute that. ( (ki)^4 = k^4 i^4 ). I know that ( i^4 = 1 ) because ( i^2 = -1 ), so ( i^4 = (i^2)^2 = (-1)^2 = 1 ). So, ( (ki)^4 = k^4 ).Second term: ( -4(ki)^3 ). Let's compute ( (ki)^3 ). That's ( k^3 i^3 ). Now, ( i^3 = i^2 cdot i = (-1)i = -i ). So, ( (ki)^3 = k^3 (-i) = -k^3 i ). Therefore, the second term becomes ( -4(-k^3 i) = 4k^3 i ).Third term: ( 6(ki)^2 ). Calculating ( (ki)^2 ), that's ( k^2 i^2 ). Since ( i^2 = -1 ), this becomes ( k^2 (-1) = -k^2 ). So, the third term is ( 6(-k^2) = -6k^2 ).Fourth term: ( -40(ki) ). That's straightforward: ( -40ki ).Fifth term: ( -64 ). That remains as it is.Putting it all together, substituting ( x = ki ) into the equation gives:( k^4 + 4k^3 i - 6k^2 - 40ki - 64 = 0 ).Now, since this equation equals zero, both the real and imaginary parts must separately equal zero. So, I can separate this into real and imaginary components.The real parts are: ( k^4 - 6k^2 - 64 ).The imaginary parts are: ( 4k^3 i - 40ki ).So, setting the imaginary part equal to zero:( 4k^3 i - 40ki = 0 ).I can factor out ( 4ki ):( 4ki(k^2 - 10) = 0 ).So, this gives me two possibilities:1. ( 4ki = 0 )2. ( k^2 - 10 = 0 )Let's consider the first possibility: ( 4ki = 0 ). Since ( i ) is not zero, this implies ( k = 0 ). But if ( k = 0 ), then ( x = 0i = 0 ). Let's check if ( x = 0 ) is a solution to the original equation.Plugging ( x = 0 ) into the equation: ( 0 - 0 + 0 - 0 - 64 = -64 neq 0 ). So, ( x = 0 ) is not a solution. Therefore, the first possibility doesn't give us a valid solution.Now, the second possibility: ( k^2 - 10 = 0 ). Solving for ( k ), we get ( k^2 = 10 ), so ( k = sqrt{10} ) or ( k = -sqrt{10} ).Therefore, the pure imaginary solutions are ( x = sqrt{10}i ) and ( x = -sqrt{10}i ).But wait, just to make sure I didn't make any mistakes, let me verify these solutions by plugging them back into the original equation.First, let's check ( x = sqrt{10}i ).Compute each term:1. ( x^4 = (sqrt{10}i)^4 = (sqrt{10})^4 (i)^4 = 100 times 1 = 100 ).2. ( -4x^3 = -4(sqrt{10}i)^3 = -4 times (10sqrt{10}i^3) = -4 times (10sqrt{10}(-i)) = 40sqrt{10}i ).3. ( 6x^2 = 6(sqrt{10}i)^2 = 6 times 10 times (-1) = -60 ).4. ( -40x = -40 times sqrt{10}i = -40sqrt{10}i ).5. ( -64 ) remains as is.Now, adding all these together:( 100 + 40sqrt{10}i - 60 - 40sqrt{10}i - 64 ).Simplify:- The ( 40sqrt{10}i ) and ( -40sqrt{10}i ) cancel each other out.- ( 100 - 60 - 64 = 100 - 124 = -24 ).Wait, that's not zero. Did I make a mistake somewhere?Let me double-check the calculations.First, ( x = sqrt{10}i ).1. ( x^4 = (sqrt{10}i)^4 = (sqrt{10})^4 times i^4 = 100 times 1 = 100 ). That seems correct.2. ( -4x^3 = -4(sqrt{10}i)^3 ). Let's compute ( (sqrt{10}i)^3 ):( (sqrt{10})^3 = 10sqrt{10} ), and ( i^3 = -i ). So, ( (sqrt{10}i)^3 = 10sqrt{10}(-i) = -10sqrt{10}i ). Therefore, ( -4x^3 = -4(-10sqrt{10}i) = 40sqrt{10}i ). That seems correct.3. ( 6x^2 = 6(sqrt{10}i)^2 ). ( (sqrt{10})^2 = 10 ), and ( i^2 = -1 ). So, ( (sqrt{10}i)^2 = 10(-1) = -10 ). Therefore, ( 6x^2 = 6(-10) = -60 ). Correct.4. ( -40x = -40 times sqrt{10}i = -40sqrt{10}i ). Correct.5. ( -64 ). Correct.Adding them up:( 100 + 40sqrt{10}i - 60 - 40sqrt{10}i - 64 ).Combine like terms:- Real parts: ( 100 - 60 - 64 = 100 - 124 = -24 ).- Imaginary parts: ( 40sqrt{10}i - 40sqrt{10}i = 0 ).So, the total is ( -24 ), not zero. Hmm, that means ( x = sqrt{10}i ) is not a solution. Did I do something wrong earlier?Wait, maybe I made a mistake when separating the real and imaginary parts. Let me go back to that step.After substitution, the equation was:( k^4 + 4k^3 i - 6k^2 - 40ki - 64 = 0 ).So, the real parts are ( k^4 - 6k^2 - 64 ), and the imaginary parts are ( 4k^3 i - 40ki ).Setting the imaginary part to zero: ( 4k^3 i - 40ki = 0 ).Factor out ( 4ki ): ( 4ki(k^2 - 10) = 0 ).So, either ( 4ki = 0 ) or ( k^2 - 10 = 0 ).As before, ( 4ki = 0 ) leads to ( k = 0 ), which isn't a solution. So, ( k^2 = 10 ), so ( k = sqrt{10} ) or ( k = -sqrt{10} ).But when I plug back ( x = sqrt{10}i ), I don't get zero. That suggests that either my substitution was wrong or my assumption that pure imaginary solutions exist is incorrect.Wait, maybe I made a mistake in the substitution. Let me try substituting ( x = ki ) again.Compute each term:1. ( x^4 = (ki)^4 = k^4 i^4 = k^4 times 1 = k^4 ).2. ( -4x^3 = -4(ki)^3 = -4k^3 i^3 = -4k^3 (-i) = 4k^3 i ).3. ( 6x^2 = 6(ki)^2 = 6k^2 i^2 = 6k^2 (-1) = -6k^2 ).4. ( -40x = -40(ki) = -40ki ).5. ( -64 ).So, putting it all together:( k^4 + 4k^3 i - 6k^2 - 40ki - 64 = 0 ).Separating real and imaginary parts:Real: ( k^4 - 6k^2 - 64 ).Imaginary: ( 4k^3 i - 40ki ).Setting imaginary part to zero:( 4k^3 i - 40ki = 0 ).Factor out ( 4ki ):( 4ki(k^2 - 10) = 0 ).So, ( k = 0 ) or ( k^2 = 10 ). As before, ( k = 0 ) doesn't work, so ( k = pm sqrt{10} ).But when I plug ( k = sqrt{10} ) into the real part:Real part: ( (sqrt{10})^4 - 6(sqrt{10})^2 - 64 = 100 - 60 - 64 = -24 neq 0 ).So, the real part isn't zero. That means that even though the imaginary part is zero, the real part isn't, so ( x = sqrt{10}i ) isn't a solution.Hmm, so does that mean there are no pure imaginary solutions? But the problem is asking to find them, so maybe I made a mistake in my approach.Wait, perhaps I need to consider that if ( x ) is purely imaginary, then the equation must hold for both real and imaginary parts. So, setting both the real and imaginary parts to zero.So, from the substitution, I have:Real part: ( k^4 - 6k^2 - 64 = 0 ).Imaginary part: ( 4k^3 - 40k = 0 ).Wait, in my earlier step, I factored out ( 4ki ), but actually, the imaginary part is ( 4k^3 i - 40ki ), which can be written as ( (4k^3 - 40k)i ). So, setting this equal to zero, we have ( 4k^3 - 40k = 0 ).So, ( 4k(k^2 - 10) = 0 ).Thus, ( k = 0 ) or ( k^2 = 10 ).Again, ( k = 0 ) doesn't work, so ( k = pm sqrt{10} ).But then, substituting ( k = sqrt{10} ) into the real part:( (sqrt{10})^4 - 6(sqrt{10})^2 - 64 = 100 - 60 - 64 = -24 neq 0 ).So, the real part isn't zero. Therefore, even though the imaginary part is zero, the real part isn't, meaning ( x = sqrt{10}i ) isn't a solution.Wait, so does that mean there are no pure imaginary solutions? But the problem is asking to find them, so maybe I made a mistake in my approach.Alternatively, perhaps I need to factor the original equation to see if it can be factored into quadratics or something, which might reveal the imaginary solutions.Let me try to factor the equation ( x^4 - 4x^3 + 6x^2 - 40x - 64 = 0 ).I notice that the coefficients resemble the expansion of ( (x - 1)^4 ), which is ( x^4 - 4x^3 + 6x^2 - 4x + 1 ). But our equation is ( x^4 - 4x^3 + 6x^2 - 40x - 64 ). So, the first three terms are the same as ( (x - 1)^4 ), but the last two terms are different.So, perhaps I can write the equation as ( (x - 1)^4 - 4x + 1 - 40x - 64 = 0 ). Wait, that might not be helpful.Alternatively, maybe I can try to factor the quartic into two quadratics.Assume ( x^4 - 4x^3 + 6x^2 - 40x - 64 = (x^2 + ax + b)(x^2 + cx + d) ).Expanding the right side:( x^4 + (a + c)x^3 + (ac + b + d)x^2 + (ad + bc)x + bd ).Set this equal to the original equation:( x^4 - 4x^3 + 6x^2 - 40x - 64 ).So, equating coefficients:1. ( a + c = -4 ) (from ( x^3 ) term).2. ( ac + b + d = 6 ) (from ( x^2 ) term).3. ( ad + bc = -40 ) (from ( x ) term).4. ( bd = -64 ) (constant term).Now, we need to find integers ( a, b, c, d ) that satisfy these equations.Looking at equation 4: ( bd = -64 ). So, possible integer pairs for ( b ) and ( d ) are:(1, -64), (-1, 64), (2, -32), (-2, 32), (4, -16), (-4, 16), (8, -8), (-8, 8), (16, -4), (-16, 4), (32, -2), (-32, 2), (64, -1), (-64, 1).Let me try some pairs and see if they work.Let's try ( b = 8 ) and ( d = -8 ). Then, equation 4 is satisfied: ( 8 times (-8) = -64 ).Now, equation 1: ( a + c = -4 ).Equation 2: ( ac + 8 + (-8) = ac = 6 ). So, ( ac = 6 ).Equation 3: ( a(-8) + c(8) = -8a + 8c = -40 ).So, from equation 1: ( c = -4 - a ).Substitute into equation 2: ( a(-4 - a) = 6 ).So, ( -4a - a^2 = 6 ).Rearranged: ( a^2 + 4a + 6 = 0 ).Discriminant: ( 16 - 24 = -8 ). No real solutions. So, this pair doesn't work.Next, try ( b = 16 ) and ( d = -4 ).Equation 4: ( 16 times (-4) = -64 ). Good.Equation 2: ( ac + 16 + (-4) = ac + 12 = 6 ). So, ( ac = -6 ).Equation 1: ( a + c = -4 ).Equation 3: ( a(-4) + c(16) = -4a + 16c = -40 ).From equation 1: ( c = -4 - a ).Substitute into equation 3:( -4a + 16(-4 - a) = -4a - 64 - 16a = -20a - 64 = -40 ).So, ( -20a = 24 ), which gives ( a = -24/20 = -6/5 ). Not an integer, so discard.Next, try ( b = -8 ) and ( d = 8 ).Equation 4: ( -8 times 8 = -64 ). Good.Equation 2: ( ac + (-8) + 8 = ac = 6 ). So, ( ac = 6 ).Equation 1: ( a + c = -4 ).Equation 3: ( a(8) + c(-8) = 8a - 8c = -40 ).From equation 1: ( c = -4 - a ).Substitute into equation 3:( 8a - 8(-4 - a) = 8a + 32 + 8a = 16a + 32 = -40 ).So, ( 16a = -72 ), ( a = -72/16 = -9/2 ). Not an integer.Next, try ( b = 4 ) and ( d = -16 ).Equation 4: ( 4 times (-16) = -64 ). Good.Equation 2: ( ac + 4 + (-16) = ac - 12 = 6 ). So, ( ac = 18 ).Equation 1: ( a + c = -4 ).Equation 3: ( a(-16) + c(4) = -16a + 4c = -40 ).From equation 1: ( c = -4 - a ).Substitute into equation 3:( -16a + 4(-4 - a) = -16a - 16 - 4a = -20a - 16 = -40 ).So, ( -20a = -24 ), ( a = 24/20 = 6/5 ). Not an integer.Next, try ( b = -4 ) and ( d = 16 ).Equation 4: ( -4 times 16 = -64 ). Good.Equation 2: ( ac + (-4) + 16 = ac + 12 = 6 ). So, ( ac = -6 ).Equation 1: ( a + c = -4 ).Equation 3: ( a(16) + c(-4) = 16a - 4c = -40 ).From equation 1: ( c = -4 - a ).Substitute into equation 3:( 16a - 4(-4 - a) = 16a + 16 + 4a = 20a + 16 = -40 ).So, ( 20a = -56 ), ( a = -56/20 = -14/5 ). Not an integer.Next, try ( b = 32 ) and ( d = -2 ).Equation 4: ( 32 times (-2) = -64 ). Good.Equation 2: ( ac + 32 + (-2) = ac + 30 = 6 ). So, ( ac = -24 ).Equation 1: ( a + c = -4 ).Equation 3: ( a(-2) + c(32) = -2a + 32c = -40 ).From equation 1: ( c = -4 - a ).Substitute into equation 3:( -2a + 32(-4 - a) = -2a - 128 - 32a = -34a - 128 = -40 ).So, ( -34a = 88 ), ( a = -88/34 = -44/17 ). Not an integer.Next, try ( b = -2 ) and ( d = 32 ).Equation 4: ( -2 times 32 = -64 ). Good.Equation 2: ( ac + (-2) + 32 = ac + 30 = 6 ). So, ( ac = -24 ).Equation 1: ( a + c = -4 ).Equation 3: ( a(32) + c(-2) = 32a - 2c = -40 ).From equation 1: ( c = -4 - a ).Substitute into equation 3:( 32a - 2(-4 - a) = 32a + 8 + 2a = 34a + 8 = -40 ).So, ( 34a = -48 ), ( a = -48/34 = -24/17 ). Not an integer.Next, try ( b = 64 ) and ( d = -1 ).Equation 4: ( 64 times (-1) = -64 ). Good.Equation 2: ( ac + 64 + (-1) = ac + 63 = 6 ). So, ( ac = -57 ).Equation 1: ( a + c = -4 ).Equation 3: ( a(-1) + c(64) = -a + 64c = -40 ).From equation 1: ( c = -4 - a ).Substitute into equation 3:( -a + 64(-4 - a) = -a - 256 - 64a = -65a - 256 = -40 ).So, ( -65a = 216 ), ( a = -216/65 ). Not an integer.Next, try ( b = -1 ) and ( d = 64 ).Equation 4: ( -1 times 64 = -64 ). Good.Equation 2: ( ac + (-1) + 64 = ac + 63 = 6 ). So, ( ac = -57 ).Equation 1: ( a + c = -4 ).Equation 3: ( a(64) + c(-1) = 64a - c = -40 ).From equation 1: ( c = -4 - a ).Substitute into equation 3:( 64a - (-4 - a) = 64a + 4 + a = 65a + 4 = -40 ).So, ( 65a = -44 ), ( a = -44/65 ). Not an integer.Hmm, this is taking too long. Maybe there's a better way. Alternatively, perhaps the equation can be factored as a product of two quadratics with real coefficients, but not necessarily integer coefficients. But that might complicate things.Alternatively, maybe I can use the rational root theorem to see if there are any rational roots, which could help in factoring.The rational root theorem says that any rational root ( p/q ) of the polynomial equation ( a_n x^n + ... + a_0 = 0 ) satisfies that ( p ) divides the constant term ( a_0 ) and ( q ) divides the leading coefficient ( a_n ).In our case, the polynomial is ( x^4 - 4x^3 + 6x^2 - 40x - 64 ). So, ( a_0 = -64 ) and ( a_n = 1 ). Therefore, any rational root must be an integer that divides -64.Possible rational roots are ( pm1, pm2, pm4, pm8, pm16, pm32, pm64 ).Let's test these:1. ( x = 1 ): ( 1 - 4 + 6 - 40 - 64 = -101 neq 0 ).2. ( x = -1 ): ( 1 + 4 + 6 + 40 - 64 = -23 neq 0 ).3. ( x = 2 ): ( 16 - 32 + 24 - 80 - 64 = -136 neq 0 ).4. ( x = -2 ): ( 16 + 32 + 24 + 80 - 64 = 96 neq 0 ).5. ( x = 4 ): ( 256 - 256 + 96 - 160 - 64 = -128 neq 0 ).6. ( x = -4 ): ( 256 + 256 + 96 + 160 - 64 = 704 neq 0 ).7. ( x = 8 ): ( 4096 - 2048 + 384 - 320 - 64 = 1048 neq 0 ).8. ( x = -8 ): ( 4096 + 2048 + 384 + 320 - 64 = 6884 neq 0 ).9. ( x = 16 ): Way too big, not zero.10. Similarly, ( x = -16 ), etc., are too big.So, no rational roots. Therefore, the polynomial doesn't factor into linear terms with integer coefficients. Maybe it factors into quadratics with real coefficients.Alternatively, perhaps I can use the substitution ( y = x - 1 ) to eliminate the cubic term. Let's try that.Let ( x = y + 1 ). Then, substitute into the equation:( (y + 1)^4 - 4(y + 1)^3 + 6(y + 1)^2 - 40(y + 1) - 64 = 0 ).Let me expand each term:1. ( (y + 1)^4 = y^4 + 4y^3 + 6y^2 + 4y + 1 ).2. ( -4(y + 1)^3 = -4(y^3 + 3y^2 + 3y + 1) = -4y^3 - 12y^2 - 12y - 4 ).3. ( 6(y + 1)^2 = 6(y^2 + 2y + 1) = 6y^2 + 12y + 6 ).4. ( -40(y + 1) = -40y - 40 ).5. ( -64 ).Now, combine all these:( y^4 + 4y^3 + 6y^2 + 4y + 1 - 4y^3 - 12y^2 - 12y - 4 + 6y^2 + 12y + 6 - 40y - 40 - 64 = 0 ).Now, combine like terms:- ( y^4 ): ( y^4 ).- ( y^3 ): ( 4y^3 - 4y^3 = 0 ).- ( y^2 ): ( 6y^2 - 12y^2 + 6y^2 = 0 ).- ( y ): ( 4y - 12y + 12y - 40y = -36y ).- Constants: ( 1 - 4 + 6 - 40 - 64 = 1 - 4 = -3; -3 + 6 = 3; 3 - 40 = -37; -37 - 64 = -101 ).So, the equation becomes:( y^4 - 36y - 101 = 0 ).Hmm, that doesn't seem much simpler. Maybe this substitution wasn't helpful.Alternatively, perhaps I can try to factor the original equation as a product of two quadratics with real coefficients, even if they aren't integers.Assume ( x^4 - 4x^3 + 6x^2 - 40x - 64 = (x^2 + ax + b)(x^2 + cx + d) ).We already tried integer coefficients, but maybe with real coefficients, it's possible.From earlier, we have:1. ( a + c = -4 ).2. ( ac + b + d = 6 ).3. ( ad + bc = -40 ).4. ( bd = -64 ).We can try to solve this system.From equation 1: ( c = -4 - a ).From equation 4: ( bd = -64 ). Let's express ( d = -64/b ).Substitute ( c = -4 - a ) and ( d = -64/b ) into equation 2:( a(-4 - a) + b + (-64/b) = 6 ).Simplify:( -4a - a^2 + b - 64/b = 6 ).Similarly, substitute into equation 3:( a(-64/b) + (-4 - a)b = -40 ).Simplify:( -64a/b - 4b - ab = -40 ).Now, we have two equations:1. ( -4a - a^2 + b - 64/b = 6 ).2. ( -64a/b - 4b - ab = -40 ).This seems complicated, but maybe we can make a substitution. Let me denote ( b = k ), then ( d = -64/k ).Then, equation 1 becomes:( -4a - a^2 + k - 64/k = 6 ).Equation 2 becomes:( -64a/k - 4k - a k = -40 ).Let me try to solve these equations.From equation 1:( -a^2 -4a + k - 64/k = 6 ).Let me rearrange:( -a^2 -4a + k - 64/k - 6 = 0 ).Multiply both sides by ( k ) to eliminate the denominator:( -a^2 k -4a k + k^2 - 64 - 6k = 0 ).Similarly, from equation 2:( -64a/k -4k - a k = -40 ).Multiply both sides by ( k ):( -64a -4k^2 - a k^2 = -40k ).Rearrange:( -64a -4k^2 - a k^2 + 40k = 0 ).Now, we have two equations:1. ( -a^2 k -4a k + k^2 - 64 - 6k = 0 ).2. ( -64a -4k^2 - a k^2 + 40k = 0 ).This is getting too complicated. Maybe there's another approach.Alternatively, perhaps the original equation can be written as ( (x^2 - 2x)^2 - 40x - 64 = 0 ).Let me check:( (x^2 - 2x)^2 = x^4 - 4x^3 + 4x^2 ).So, ( x^4 - 4x^3 + 4x^2 - 40x - 64 = 0 ).But the original equation is ( x^4 - 4x^3 + 6x^2 - 40x - 64 = 0 ). So, the difference is in the ( x^2 ) term: 6x^2 vs. 4x^2. So, ( (x^2 - 2x)^2 + 2x^2 - 40x - 64 = 0 ).Hmm, not helpful.Alternatively, maybe complete the square for the quartic.But quartic equations are complicated. Maybe I can use the substitution ( y = x^2 ), but that might not help since the equation has odd powers.Wait, another idea: since we're looking for pure imaginary solutions, maybe the equation can be transformed into a quadratic in terms of ( y = x^2 ), but considering ( x = ki ), so ( y = (ki)^2 = -k^2 ).Let me try that.Let ( x = ki ), so ( x^2 = -k^2 ), ( x^3 = -k^3 i ), ( x^4 = k^4 ).Substitute into the equation:( k^4 - 4(-k^3 i) + 6(-k^2) - 40(ki) - 64 = 0 ).Simplify:( k^4 + 4k^3 i - 6k^2 - 40ki - 64 = 0 ).Which is the same as before. So, separating real and imaginary parts:Real: ( k^4 - 6k^2 - 64 = 0 ).Imaginary: ( 4k^3 - 40k = 0 ).From the imaginary part, we have ( 4k(k^2 - 10) = 0 ), so ( k = 0 ) or ( k = pm sqrt{10} ).But as before, substituting ( k = sqrt{10} ) into the real part gives ( 100 - 60 - 64 = -24 neq 0 ). So, no solution.Wait, so does that mean there are no pure imaginary solutions? But the problem is asking to find them, so maybe I made a mistake in my approach.Alternatively, perhaps I need to consider that the equation might have complex solutions, but not purely imaginary. But the problem specifically asks for pure imaginary solutions.Wait, maybe I need to check my substitution again.Let me try substituting ( x = ki ) into the original equation:( (ki)^4 - 4(ki)^3 + 6(ki)^2 - 40(ki) - 64 = 0 ).Compute each term:1. ( (ki)^4 = k^4 i^4 = k^4 times 1 = k^4 ).2. ( -4(ki)^3 = -4k^3 i^3 = -4k^3 (-i) = 4k^3 i ).3. ( 6(ki)^2 = 6k^2 i^2 = 6k^2 (-1) = -6k^2 ).4. ( -40(ki) = -40ki ).5. ( -64 ).So, the equation becomes:( k^4 + 4k^3 i - 6k^2 - 40ki - 64 = 0 ).Separate into real and imaginary parts:Real: ( k^4 - 6k^2 - 64 ).Imaginary: ( 4k^3 - 40k ).Set both to zero:1. ( 4k^3 - 40k = 0 ) → ( 4k(k^2 - 10) = 0 ) → ( k = 0 ) or ( k = pm sqrt{10} ).2. ( k^4 - 6k^2 - 64 = 0 ).Now, let's solve the real part equation: ( k^4 - 6k^2 - 64 = 0 ).Let me set ( m = k^2 ), so the equation becomes:( m^2 - 6m - 64 = 0 ).Solve for ( m ):( m = [6 pm sqrt{36 + 256}]/2 = [6 pm sqrt{292}]/2 = [6 pm 2sqrt{73}]/2 = 3 pm sqrt{73} ).So, ( k^2 = 3 + sqrt{73} ) or ( k^2 = 3 - sqrt{73} ).But ( 3 - sqrt{73} ) is negative because ( sqrt{73} approx 8.544 ), so ( 3 - 8.544 approx -5.544 ). Therefore, ( k^2 ) can't be negative, so only ( k^2 = 3 + sqrt{73} ).Thus, ( k = pm sqrt{3 + sqrt{73}} ).But earlier, from the imaginary part, we had ( k = pm sqrt{10} ). So, unless ( sqrt{10} = sqrt{3 + sqrt{73}} ), which is not true because ( 3 + sqrt{73} approx 3 + 8.544 = 11.544 ), and ( sqrt{11.544} approx 3.398 ), while ( sqrt{10} approx 3.162 ). So, they are not equal.Therefore, there is no value of ( k ) that satisfies both the real and imaginary parts being zero. Hence, there are no pure imaginary solutions to the equation.Wait, but the problem is asking to find the pure imaginary solutions. So, maybe I made a mistake in my earlier assumption that pure imaginary solutions exist. Alternatively, perhaps the problem has a typo, or I misread it.Alternatively, maybe I need to consider that the equation can be factored in a way that reveals the imaginary solutions. Let me try to factor the original equation again.Wait, another idea: perhaps the equation can be written as ( (x^2 - 2x)^2 - 40x - 64 = 0 ).Compute ( (x^2 - 2x)^2 = x^4 - 4x^3 + 4x^2 ).So, the equation becomes:( x^4 - 4x^3 + 4x^2 - 40x - 64 = 0 ).But the original equation is ( x^4 - 4x^3 + 6x^2 - 40x - 64 = 0 ). So, the difference is in the ( x^2 ) term: 6x^2 vs. 4x^2. So, ( (x^2 - 2x)^2 + 2x^2 - 40x - 64 = 0 ).Not helpful.Alternatively, maybe I can use the fact that if ( x ) is purely imaginary, then ( x = ki ), and substitute into the equation, but also consider that the equation must hold for both real and imaginary parts. So, setting both parts to zero.From earlier, we have:Real part: ( k^4 - 6k^2 - 64 = 0 ).Imaginary part: ( 4k^3 - 40k = 0 ).So, solving the imaginary part gives ( k = 0 ) or ( k = pm sqrt{10} ).But substituting ( k = sqrt{10} ) into the real part gives ( 100 - 60 - 64 = -24 neq 0 ).Therefore, there are no pure imaginary solutions because the real part doesn't equal zero for the values of ( k ) that make the imaginary part zero.Wait, but the problem is asking to find the pure imaginary solutions. So, maybe the answer is that there are no pure imaginary solutions.But in the initial attempt, I thought ( x = sqrt{10}i ) and ( x = -sqrt{10}i ) were solutions, but when I checked, they didn't satisfy the equation. So, perhaps the correct answer is that there are no pure imaginary solutions.Alternatively, maybe I made a mistake in the substitution. Let me double-check.Substituting ( x = sqrt{10}i ):1. ( x^4 = (sqrt{10}i)^4 = (10)^2 (i^4) = 100 times 1 = 100 ).2. ( -4x^3 = -4(sqrt{10}i)^3 = -4 times (10sqrt{10}i^3) = -4 times (10sqrt{10}(-i)) = 40sqrt{10}i ).3. ( 6x^2 = 6(sqrt{10}i)^2 = 6 times 10 times (-1) = -60 ).4. ( -40x = -40 times sqrt{10}i = -40sqrt{10}i ).5. ( -64 ).Adding them up:( 100 + 40sqrt{10}i - 60 - 40sqrt{10}i - 64 = 100 - 60 - 64 + (40sqrt{10}i - 40sqrt{10}i) = -24 + 0i neq 0 ).So, indeed, ( x = sqrt{10}i ) is not a solution.Therefore, the conclusion is that there are no pure imaginary solutions to the equation.But the problem is asking to find them, so maybe I made a mistake in my approach. Alternatively, perhaps the problem is designed to have no pure imaginary solutions, and the answer is that there are none.Wait, but in the initial substitution, I thought ( k = pm sqrt{10} ) would work, but they don't. So, perhaps the answer is that there are no pure imaginary solutions.Alternatively, maybe I need to consider that the equation can be factored into quadratics with complex coefficients, but that's beyond the scope here.Given all this, I think the correct answer is that there are no pure imaginary solutions to the equation.

💡First, I identify the total weight of oranges, which is 1035 pounds, and the weight of oranges in each bag, which is 23 pounds.To find the number of bags, I divide the total weight by the weight per bag: 1035 ÷ 23.Performing the division gives me 45.Therefore, there are 45 bags of oranges.

💡To compare the sizes of (-frac{5}{6}) and (-frac{7}{9}), we first convert them to have a common denominator for easier comparison:1. Convert (-frac{5}{6}) and (-frac{7}{9}) to have a common denominator: - (left|-frac{5}{6}right| = frac{5}{6} = frac{5 times 6}{6 times 6} = frac{30}{36}) - (left|-frac{7}{9}right| = frac{7}{9} = frac{7 times 4}{9 times 4} = frac{28}{36})2. Compare the absolute values: - Since (frac{30}{36} > frac{28}{36}), it means the absolute value of (-frac{5}{6}) is greater than the absolute value of (-frac{7}{9}).3. Determine the relationship based on the original negative values: - Because a larger absolute value corresponds to a smaller value when both numbers are negative, (-frac{5}{6} < -frac{7}{9}).Therefore, the correct comparison symbol to fill in the blank is (<), encapsulating the final answer as (boxed{<}).

💡Okay, so I have this triangle ABC, and I know angle A is 30 degrees, and the side opposite to angle A, which is side a, is equal to the square root of 3. The question is asking me to find the value of (a + b + c) divided by (sin A + sin B + sin C). Hmm, that seems a bit complicated, but maybe I can break it down.First, I remember that in any triangle, the Law of Sines relates the sides and angles. It says that a divided by sin A is equal to b divided by sin B, which is equal to c divided by sin C, and all of these are equal to twice the radius of the circumcircle of the triangle. So, mathematically, that's written as:a / sin A = b / sin B = c / sin C = 2RWhere R is the radius of the circumcircle. Maybe I can use this to find some relationships between the sides and angles.Given that angle A is 30 degrees and side a is sqrt(3), let me plug those into the Law of Sines. So, a / sin A should be equal to 2R. Let me calculate that:a / sin A = sqrt(3) / sin(30°)I know that sin(30°) is 0.5, so:sqrt(3) / 0.5 = 2 * sqrt(3)So, 2R = 2 * sqrt(3), which means R = sqrt(3). Okay, so the circumradius R is sqrt(3).Now, since all sides are related to the sine of their opposite angles through 2R, I can express sides b and c in terms of angles B and C. So, using the same Law of Sines:b = 2R * sin B = 2 * sqrt(3) * sin Bc = 2R * sin C = 2 * sqrt(3) * sin CSo, now I can write expressions for sides b and c in terms of angles B and C.Now, the problem is asking for (a + b + c) / (sin A + sin B + sin C). Let me write that out:(a + b + c) / (sin A + sin B + sin C)I already know a is sqrt(3), and I have expressions for b and c in terms of sin B and sin C. Let me substitute those in:(sqrt(3) + 2 * sqrt(3) * sin B + 2 * sqrt(3) * sin C) / (sin 30° + sin B + sin C)Hmm, let me factor out the 2 * sqrt(3) from the numerator:sqrt(3) + 2 * sqrt(3) * (sin B + sin C) all over (sin 30° + sin B + sin C)Wait, that's not quite right. Let me see: sqrt(3) is just a constant, and then 2 * sqrt(3) times (sin B + sin C). So, the numerator is sqrt(3) + 2 * sqrt(3) * (sin B + sin C). The denominator is sin 30° + sin B + sin C.I can factor sqrt(3) out of the numerator:sqrt(3) * [1 + 2(sin B + sin C)] / [sin 30° + sin B + sin C]But I know that sin 30° is 0.5, so the denominator becomes 0.5 + sin B + sin C. Let me write that:sqrt(3) * [1 + 2(sin B + sin C)] / [0.5 + sin B + sin C]Hmm, maybe I can simplify this expression further. Let me denote S = sin B + sin C for simplicity. Then the numerator becomes sqrt(3) * [1 + 2S] and the denominator becomes 0.5 + S.So, the expression is sqrt(3) * (1 + 2S) / (0.5 + S)Let me see if I can factor or simplify this. Let me write 1 + 2S as 2S + 1, and 0.5 + S as S + 0.5. Maybe I can factor something out.Wait, 1 + 2S is equal to 2*(0.5 + S) + 0? Let me check:2*(0.5 + S) = 1 + 2S. Oh, that's exactly the numerator! So, 1 + 2S = 2*(0.5 + S). Therefore, the numerator becomes sqrt(3) * 2*(0.5 + S), and the denominator is (0.5 + S).So, sqrt(3) * 2*(0.5 + S) / (0.5 + S) = sqrt(3) * 2, because (0.5 + S) cancels out.So, that simplifies to 2 * sqrt(3). Therefore, the value of (a + b + c)/(sin A + sin B + sin C) is 2 * sqrt(3).Wait, let me check if I did everything correctly. I used the Law of Sines to express b and c in terms of sin B and sin C, then substituted into the expression, factored, and simplified. It seems correct. So, the answer should be 2 * sqrt(3).

💡Okay, so I have this problem here about a line in the coordinate plane. The line is given by the equation y = kx + 3. It intersects the x-axis at point A and the y-axis at point B. Then, the line is shifted 5 units to the right, and the region swept by the line segment AB forms a rhombus. I need to find the value of k.Alright, let me break this down step by step. First, I should probably find the coordinates of points A and B. Since A is where the line intersects the x-axis, that means y = 0 at point A. So I can set y to 0 in the equation and solve for x.So, 0 = kx + 3. That means kx = -3, so x = -3/k. Therefore, point A is at (-3/k, 0). Next, point B is where the line intersects the y-axis. That happens when x = 0. Plugging x = 0 into the equation, y = k*0 + 3, so y = 3. So point B is at (0, 3).Now, the line is shifted 5 units to the right. Shifting a line to the right by 5 units means replacing x with (x - 5) in the equation. So the new equation becomes y = k(x - 5) + 3, which simplifies to y = kx - 5k + 3.This shifted line will intersect the x-axis and y-axis at new points, let's call them A' and B'. Let me find those points.For point A', set y = 0 in the shifted equation: 0 = kx - 5k + 3. Solving for x, we get kx = 5k - 3, so x = (5k - 3)/k. Therefore, point A' is at ((5k - 3)/k, 0).For point B', set x = 0 in the shifted equation: y = k*0 - 5k + 3, so y = -5k + 3. Therefore, point B' is at (0, -5k + 3).Now, the region swept by the line segment AB when the line is shifted forms a rhombus. A rhombus is a quadrilateral with all sides equal in length, and opposite sides are parallel. The diagonals of a rhombus bisect each other at right angles.So, the original segment AB and the shifted segment A'B' are two sides of the rhombus. The other two sides are the paths traced by points A and B as the line shifts from AB to A'B'. Wait, actually, when the line shifts, every point on the line moves 5 units to the right. So, the segment AB moves from its original position to A'B', sweeping out a region. The boundary of this region is a rhombus. So, the rhombus is formed by the original segment AB, the shifted segment A'B', and the two lines connecting A to A' and B to B'. Therefore, the rhombus has vertices at A, B, A', and B'. So, the four vertices of the rhombus are A(-3/k, 0), B(0, 3), A'((5k - 3)/k, 0), and B'(0, -5k + 3).Now, since it's a rhombus, all sides must be equal. So, the lengths of AB, BA', A'B', and B'A must be equal.Alternatively, since the sides are formed by the movement of the line, the sides of the rhombus are the original segment AB, the shifted segment A'B', and the two connecting segments AA' and BB'.But perhaps another approach is to consider the vectors involved. The shift is 5 units to the right, so the vector from A to A' is (5, 0), and similarly, the vector from B to B' is (5, 0). Wait, but in reality, when you shift a line, every point on the line moves 5 units to the right. So, point A moves from (-3/k, 0) to ((-3/k) + 5, 0), which is (5 - 3/k, 0). Similarly, point B moves from (0, 3) to (5, 3). Wait, hold on, that contradicts my earlier calculation. Let me check.If the original line is y = kx + 3, and we shift it 5 units to the right, the new line is y = k(x - 5) + 3. So, to find the new x-intercept, set y = 0:0 = k(x - 5) + 3 => kx - 5k + 3 = 0 => kx = 5k - 3 => x = (5k - 3)/k.So, point A' is ((5k - 3)/k, 0). Similarly, the y-intercept is when x = 0: y = k(0 - 5) + 3 = -5k + 3. So, point B' is (0, -5k + 3).But when I think about shifting the entire line 5 units to the right, each point on the line should move 5 units to the right. So, point A(-3/k, 0) should move to (-3/k + 5, 0), which is (5 - 3/k, 0). Similarly, point B(0, 3) should move to (5, 3). Wait, so there's a discrepancy here. Which one is correct?I think the confusion arises from how the shift affects the equation versus how it affects individual points. When you shift a line 5 units to the right, the equation becomes y = k(x - 5) + 3, which is correct. So, the new x-intercept is (5k - 3)/k, and the new y-intercept is -5k + 3.However, if you consider shifting each point on the line 5 units to the right, then point A(-3/k, 0) would move to (5 - 3/k, 0), and point B(0, 3) would move to (5, 3). But these two results don't match. So, which one is correct?I think the issue is that shifting the line 5 units to the right affects the equation, which in turn affects the intercepts. So, the intercepts are not simply shifted by 5 units; instead, the entire line is transformed, which changes where it crosses the axes.Therefore, the correct new intercepts are A'((5k - 3)/k, 0) and B'(0, -5k + 3).So, the original segment AB connects (-3/k, 0) to (0, 3), and the shifted segment A'B' connects ((5k - 3)/k, 0) to (0, -5k + 3).Now, the region swept by AB as it shifts to A'B' is a rhombus. So, the four sides of the rhombus are AB, A'B', and the two lines connecting A to A' and B to B'.Therefore, the sides of the rhombus are AB, A'B', AA', and BB'.Since it's a rhombus, all four sides must be equal in length.So, let's compute the lengths of AB, A'B', AA', and BB' and set them equal.First, compute AB:Points A(-3/k, 0) and B(0, 3).Distance AB = sqrt[(0 - (-3/k))^2 + (3 - 0)^2] = sqrt[(3/k)^2 + 9] = sqrt(9/k² + 9).Similarly, compute A'B':Points A'((5k - 3)/k, 0) and B'(0, -5k + 3).Distance A'B' = sqrt[(0 - (5k - 3)/k)^2 + (-5k + 3 - 0)^2] = sqrt[( (3 - 5k)/k )^2 + (-5k + 3)^2].Simplify:First term: (3 - 5k)^2 / k²Second term: (-5k + 3)^2 = (5k - 3)^2So, A'B' = sqrt[ ( (3 - 5k)^2 ) / k² + (5k - 3)^2 ]Note that (3 - 5k)^2 = (5k - 3)^2, so we can write:A'B' = sqrt[ ( (5k - 3)^2 ) / k² + (5k - 3)^2 ]Factor out (5k - 3)^2:A'B' = sqrt[ (5k - 3)^2 (1/k² + 1) ) ] = |5k - 3| * sqrt(1/k² + 1)Similarly, AB = sqrt(9/k² + 9) = sqrt(9(1/k² + 1)) = 3 sqrt(1/k² + 1)Since AB and A'B' are sides of the rhombus, they must be equal:AB = A'B'So,3 sqrt(1/k² + 1) = |5k - 3| sqrt(1/k² + 1)Assuming sqrt(1/k² + 1) ≠ 0, which it isn't because k is a real number (except k=0, but k=0 would make the line horizontal, and shifting it wouldn't form a rhombus as the segment AB would remain horizontal). So, we can divide both sides by sqrt(1/k² + 1):3 = |5k - 3|So, |5k - 3| = 3This gives two cases:Case 1: 5k - 3 = 3 => 5k = 6 => k = 6/5Case 2: 5k - 3 = -3 => 5k = 0 => k = 0But k=0 would make the line y=3, which is horizontal. Shifting it 5 units to the right would result in y=3, which is the same line. So, the region swept would not be a rhombus but just the line itself. Therefore, k=0 is not a valid solution.Therefore, k = 6/5.Wait, but let me check if this makes sense.If k = 6/5, then the original line is y = (6/5)x + 3.Point A is (-3/(6/5), 0) = (-3*(5/6), 0) = (-5/2, 0).Point B is (0, 3).Shifted line is y = (6/5)(x - 5) + 3 = (6/5)x - 6 + 3 = (6/5)x - 3.So, point A' is where y=0: 0 = (6/5)x - 3 => (6/5)x = 3 => x = (3)*(5/6) = 5/2. So, A' is (5/2, 0).Point B' is where x=0: y = (6/5)(-5) + 3 = -6 + 3 = -3. So, B' is (0, -3).So, the four points of the rhombus are A(-5/2, 0), B(0, 3), A'(5/2, 0), and B'(0, -3).Let me plot these points mentally:- A is (-2.5, 0)- B is (0, 3)- A' is (2.5, 0)- B' is (0, -3)Connecting these points, the figure should be a rhombus.Let me compute the lengths of the sides:AB: distance from (-2.5, 0) to (0, 3):sqrt[(2.5)^2 + 3^2] = sqrt[6.25 + 9] = sqrt[15.25] ≈ 3.905A'B': distance from (2.5, 0) to (0, -3):sqrt[(2.5)^2 + (-3)^2] = sqrt[6.25 + 9] = sqrt[15.25] ≈ 3.905AA': distance from (-2.5, 0) to (2.5, 0):sqrt[(5)^2 + 0^2] = 5BB': distance from (0, 3) to (0, -3):sqrt[0^2 + (-6)^2] = 6Wait, but in a rhombus, all sides should be equal. However, here AA' is 5 and BB' is 6, which are not equal. So, something is wrong.Hmm, that suggests that my earlier conclusion that k=6/5 is incorrect.Wait, perhaps I made a mistake in assuming that AB and A'B' are sides of the rhombus. Maybe the sides of the rhombus are not AB and A'B', but rather the segments connecting A to A' and B to B', and the original and shifted segments.Wait, let me think again.When the line shifts from AB to A'B', the region swept is a parallelogram, specifically a rhombus. The sides of the rhombus are the original segment AB, the shifted segment A'B', and the two lines connecting A to A' and B to B'.But in a rhombus, all four sides are equal. So, AB must equal A'B', and AA' must equal BB'.But in my calculation, AB = A'B' ≈ 3.905, AA' = 5, BB' = 6. So, they are not equal. Therefore, k=6/5 does not satisfy the condition.So, my earlier approach must be flawed.Perhaps I need to consider the diagonals of the rhombus instead.In a rhombus, the diagonals are perpendicular bisectors of each other. The diagonals are AB and A'B', or maybe AA' and BB'.Wait, actually, when you shift the line, the original segment AB and the shifted segment A'B' are two sides of the rhombus, but they are not the diagonals.Wait, maybe the diagonals are AA' and BB'.Because when you shift the line, the points A and A' are connected, and B and B' are connected, forming the other two sides.But in a rhombus, the diagonals are not necessarily equal, but they bisect each other at right angles.Alternatively, perhaps the diagonals are AB and A'B', but that seems unlikely.Wait, perhaps I need to think differently.The region swept by AB is a rhombus, which means that the movement of AB from its original position to the shifted position forms a rhombus. So, the sides of the rhombus are the original AB, the shifted A'B', and the two lines connecting A to A' and B to B'.Therefore, the four sides are AB, A'B', AA', and BB'.For it to be a rhombus, all four sides must be equal in length.So, AB = A'B' = AA' = BB'So, AB = A'B' and AA' = BB'Earlier, I found that AB = 3 sqrt(1/k² + 1) and A'B' = |5k - 3| sqrt(1/k² + 1). Setting AB = A'B' gives |5k - 3| = 3, leading to k=6/5 or k=0, but k=0 is invalid.But then, AA' is the distance from A to A', which is 5 units, since shifting 5 units to the right. Similarly, BB' is the distance from B to B', which is sqrt[(5)^2 + (-5k)^2] because B moves from (0,3) to (5, -5k + 3). Wait, no, actually, when shifting the line, each point moves 5 units to the right, so the vertical component remains the same? Wait, no, shifting a line 5 units to the right affects both x and y coordinates depending on the slope.Wait, perhaps I need to clarify how points A and B move when the line is shifted.When you shift the entire line 5 units to the right, every point on the line moves 5 units to the right. So, point A(-3/k, 0) moves to (-3/k + 5, 0), and point B(0, 3) moves to (5, 3). But earlier, when I shifted the equation, I found that the new x-intercept is (5k - 3)/k and the new y-intercept is -5k + 3. So, which is correct?I think the confusion comes from two different interpretations of "shifting the line." If you shift the line 5 units to the right, it's equivalent to replacing x with x - 5 in the equation, which changes the intercepts as I calculated earlier. However, if you consider shifting each point on the line 5 units to the right, then the intercepts would simply increase by 5 in the x-direction.But these two interpretations lead to different results. So, which one is correct?I think the correct interpretation is that shifting the line 5 units to the right is equivalent to replacing x with x - 5 in the equation, which changes the intercepts accordingly. Therefore, the new intercepts are A'((5k - 3)/k, 0) and B'(0, -5k + 3).Therefore, the movement of point A is from (-3/k, 0) to ((5k - 3)/k, 0), which is a horizontal shift of 5 units to the right, as (5k - 3)/k = 5 - 3/k, so the x-coordinate increases by 5.Similarly, point B moves from (0, 3) to (0, -5k + 3). Wait, but that's not a horizontal shift; that's a vertical shift. That doesn't make sense because shifting the line to the right should only affect the x-coordinates, not the y-coordinates.Wait, no, actually, when you shift the line to the right, the y-intercept changes because the line's position relative to the y-axis changes. So, the y-intercept is not just shifted right; it's recalculated based on the new equation.Therefore, point B moves from (0, 3) to (0, -5k + 3). So, the movement is not purely horizontal; it's a combination of horizontal and vertical movement because the line is tilted.Therefore, the distance from B to B' is not just 5 units, but the actual distance between (0, 3) and (0, -5k + 3), which is |3 - (-5k + 3)| = |5k|.Similarly, the distance from A to A' is |(-3/k) - ((5k - 3)/k)| = |(-3/k) - 5 + 3/k| = |-5| = 5.So, AA' is 5 units, and BB' is |5k| units.Since in the rhombus, all sides must be equal, so AA' = BB', which gives |5k| = 5 => |k| = 1 => k = ±1.But earlier, we had from AB = A'B', |5k - 3| = 3, leading to k = 6/5 or k=0.So, now we have two conditions:1. |5k - 3| = 3 => k = 6/5 or k=0 (invalid)2. |5k| = 5 => |k| = 1 => k = ±1But these two conditions must both be satisfied for the rhombus to form. However, k=6/5 and k=±1 are different. Therefore, there is no solution that satisfies both conditions unless 6/5 = ±1, which is not true.This suggests that my approach is incorrect.Perhaps I need to consider that the sides of the rhombus are not AB, A'B', AA', and BB', but rather the segments AB, A'B', and the two lines connecting A to B' and B to A'.Wait, that might make more sense. Because when you shift the line, the original segment AB sweeps out a region, and the new segment A'B' is connected to the original segment via the movement of the line.So, the rhombus would have vertices at A, B, B', A'.Wait, let me visualize this. The original segment AB is from A(-3/k, 0) to B(0, 3). The shifted segment A'B' is from A'((5k - 3)/k, 0) to B'(0, -5k + 3). The region swept by AB is the area between AB and A'B', connected by the lines AA' and BB'.But if the figure is a rhombus, then the sides must be AB, A'B', and the two connecting lines AA' and BB', all equal in length.But as I saw earlier, AB = A'B' ≈ 3.905 when k=6/5, but AA' = 5 and BB' = 6, which are not equal. Therefore, k=6/5 is not a solution.Alternatively, if I consider the sides to be AB, BB', A'B', and AA', but that also doesn't resolve the issue.Wait, perhaps I need to think about the vectors involved. The shift is a translation vector of (5, 0). So, the movement from A to A' is (5, 0), and from B to B' is also (5, 0). But in reality, the movement from B to B' is not purely horizontal because the line is tilted.Wait, no, the translation vector is (5, 0), so every point on the line moves 5 units to the right. Therefore, point A moves from (-3/k, 0) to (-3/k + 5, 0), and point B moves from (0, 3) to (5, 3). But earlier, when I shifted the equation, I found that the new y-intercept is -5k + 3, which suggests that point B moves to (0, -5k + 3). This is conflicting.I think the confusion arises because shifting the line 5 units to the right changes the equation, which in turn changes the intercepts. However, if we consider the line as a set of points, each point on the line is shifted 5 units to the right, so point B(0, 3) would move to (5, 3), not to (0, -5k + 3).Therefore, perhaps my earlier calculation of A' and B' was incorrect.Let me clarify:If the line is shifted 5 units to the right, then every point (x, y) on the original line is moved to (x + 5, y). Therefore, the new line is the set of points (x + 5, y) where (x, y) is on the original line y = kx + 3. Therefore, the new line can be expressed as y = k(x - 5) + 3, which is the same as shifting the equation.However, the intercepts of the new line are calculated by setting y=0 and x=0:For y=0: 0 = k(x - 5) + 3 => kx - 5k + 3 = 0 => x = (5k - 3)/k.For x=0: y = k(-5) + 3 = -5k + 3.So, point A' is ((5k - 3)/k, 0), and point B' is (0, -5k + 3).But if we consider shifting each point 5 units to the right, point A(-3/k, 0) moves to (-3/k + 5, 0), and point B(0, 3) moves to (5, 3).Therefore, there are two different interpretations:1. Shifting the line 5 units to the right, which changes the equation and results in new intercepts A'((5k - 3)/k, 0) and B'(0, -5k + 3).2. Shifting each point on the line 5 units to the right, resulting in A'(-3/k + 5, 0) and B'(5, 3).These are two different scenarios. The problem states that the line is shifted 5 units to the right, which typically means the first interpretation, changing the equation. However, the region swept by the segment AB would involve the movement of the entire segment, which would be the second interpretation.Therefore, perhaps the correct approach is to consider that the segment AB is moved 5 units to the right, resulting in segment A'B', where A' is (-3/k + 5, 0) and B' is (5, 3). Then, the region swept by AB is the area between AB and A'B', connected by the lines AA' and BB'.In this case, the four vertices of the rhombus would be A(-3/k, 0), B(0, 3), B'(5, 3), and A'(-3/k + 5, 0).Wait, but connecting these points, the figure would have sides AB, BB', B'A', and A'A.But let me compute the lengths:AB: distance from A to B = sqrt[(3/k)^2 + 9]BB': distance from B(0,3) to B'(5,3) = 5 unitsB'A': distance from B'(5,3) to A'(-3/k + 5, 0) = sqrt[( (-3/k)^2 + 9 ]A'A: distance from A'(-3/k + 5, 0) to A(-3/k, 0) = 5 unitsSo, sides AB and B'A' have the same length, and sides BB' and A'A are both 5 units.Therefore, the figure is a parallelogram with opposite sides equal. For it to be a rhombus, all four sides must be equal. Therefore, AB must equal BB', which is 5 units.So, AB = 5.Compute AB:AB = sqrt[(3/k)^2 + 9] = 5So,sqrt(9/k² + 9) = 5Square both sides:9/k² + 9 = 25Subtract 9:9/k² = 16Multiply both sides by k²:9 = 16k²Divide by 16:k² = 9/16Take square root:k = ±3/4Therefore, the value of k is ±3/4.Let me verify this.If k = 3/4, then the original line is y = (3/4)x + 3.Point A is (-3/(3/4), 0) = (-4, 0).Point B is (0, 3).Shifting the line 5 units to the right, the new line is y = (3/4)(x - 5) + 3 = (3/4)x - 15/4 + 3 = (3/4)x - 15/4 + 12/4 = (3/4)x - 3/4.So, the new x-intercept is when y=0:0 = (3/4)x - 3/4 => (3/4)x = 3/4 => x = 1. So, A' is (1, 0).The new y-intercept is when x=0: y = -3/4. So, B' is (0, -3/4).Wait, but according to the earlier interpretation, shifting each point 5 units to the right, point A(-4, 0) would move to (1, 0), and point B(0, 3) would move to (5, 3). So, A' is (1, 0) and B' is (5, 3).Therefore, the four points of the rhombus are A(-4, 0), B(0, 3), B'(5, 3), and A'(1, 0).Wait, but connecting these points, the sides are AB, BB', B'A', and A'A.Compute the lengths:AB: distance from (-4,0) to (0,3): sqrt[4² + 3²] = 5BB': distance from (0,3) to (5,3): 5B'A': distance from (5,3) to (1,0): sqrt[(5-1)^2 + (3-0)^2] = sqrt[16 + 9] = 5A'A: distance from (1,0) to (-4,0): 5So, all sides are 5 units, which is a rhombus.Similarly, if k = -3/4, the original line is y = (-3/4)x + 3.Point A is (-3/(-3/4), 0) = (4, 0).Point B is (0, 3).Shifting the line 5 units to the right, the new line is y = (-3/4)(x - 5) + 3 = (-3/4)x + 15/4 + 3 = (-3/4)x + 15/4 + 12/4 = (-3/4)x + 27/4.So, the new x-intercept is when y=0:0 = (-3/4)x + 27/4 => (3/4)x = 27/4 => x = 9. So, A' is (9, 0).The new y-intercept is when x=0: y = 27/4. So, B' is (0, 27/4).But according to shifting each point 5 units to the right, point A(4, 0) moves to (9, 0), and point B(0, 3) moves to (5, 3).Therefore, the four points of the rhombus are A(4, 0), B(0, 3), B'(5, 3), and A'(9, 0).Compute the lengths:AB: distance from (4,0) to (0,3): sqrt[4² + 3²] = 5BB': distance from (0,3) to (5,3): 5B'A': distance from (5,3) to (9,0): sqrt[(9-5)^2 + (0-3)^2] = sqrt[16 + 9] = 5A'A: distance from (9,0) to (4,0): 5Again, all sides are 5 units, forming a rhombus.Therefore, the correct values of k are ±3/4.

💡Okay, so I have this problem here with four statements about the symmetry of function graphs. I need to figure out how many of these statements are correct. Let me go through each one step by step.Starting with statement ①: "The graph of the function ( y = f(x) ) is symmetric to the graph of the function ( y = f(-x) ) with respect to the line ( x = 0 )." Hmm, I remember that reflecting a function over the y-axis (which is the line ( x = 0 )) involves replacing ( x ) with ( -x ). So, if you have ( f(x) ) and ( f(-x) ), their graphs should be mirror images across the y-axis. That makes sense. So, statement ① seems correct.Moving on to statement ②: "The graph of the function ( y = f(x) ) is symmetric to the graph of the function ( y = -f(x) ) with respect to the line ( y = 0 )." Reflecting over the x-axis (which is ( y = 0 )) involves negating the function, so ( f(x) ) becomes ( -f(x) ). That should make their graphs symmetric across the x-axis. So, statement ② also seems correct.Now, statement ③: "The graph of the function ( y = f(x) ) is symmetric to the graph of the function ( y = -f(-x) ) with respect to the origin." Reflecting over the origin involves both negating the x and y coordinates. So, if you take ( f(x) ) and replace ( x ) with ( -x ) and ( y ) with ( -y ), you get ( -y = f(-x) ) or ( y = -f(-x) ). That sounds right. So, statement ③ is correct too.Finally, statement ④: "If for all ( x in mathbb{R} ), the function ( y = f(x) ) satisfies ( f(a + x) = f(a - x) ), then the graph of ( y = f(x) ) is symmetric with respect to the line ( x = a )." I recall that symmetry about a vertical line ( x = a ) means that for any point ( (a + x, y) ) on the graph, there is a corresponding point ( (a - x, y) ). This condition ( f(a + x) = f(a - x) ) exactly captures that idea. So, statement ④ is correct as well.Wait, so all four statements seem correct? But let me double-check each one to make sure I didn't make a mistake.For statement ①, reflecting over the y-axis does give ( f(-x) ), so that's correct.Statement ②, reflecting over the x-axis gives ( -f(x) ), which is also correct.Statement ③, reflecting over the origin involves both negating x and y, leading to ( -f(-x) ), so that's correct too.And statement ④, the condition ( f(a + x) = f(a - x) ) is the definition of symmetry about ( x = a ), so that's correct as well.Hmm, so all four statements are correct. But I feel like sometimes these multiple-choice questions might have a trick, but I don't see it here. Each statement aligns with the definitions of symmetry I know. So, I think all four are correct.**Final Answer**boxed{D}