💡First, I recognize that the total probability of all possible outcomes must equal 1. The spinner has four sections labeled A, B, C, and D, so the sum of their probabilities should be 1.I know the probabilities for A, B, and D:- P(A) = 3/8- P(B) = 1/4- P(D) = 1/8To find P(C), I can set up the equation:[ P(A) + P(B) + P(C) + P(D) = 1 ]Substituting the known values:[ frac{3}{8} + frac{1}{4} + P(C) + frac{1}{8} = 1 ]Next, I'll convert all fractions to have a common denominator to simplify the calculation. The common denominator for 8 and 4 is 8:[ frac{3}{8} + frac{2}{8} + P(C) + frac{1}{8} = 1 ]Now, I'll add the fractions:[ frac{6}{8} + P(C) = 1 ]To isolate P(C), I'll subtract 6/8 from both sides of the equation:[ P(C) = 1 - frac{6}{8} ]Simplifying the right side:[ P(C) = frac{2}{8} = frac{1}{4} ]Therefore, the probability of the spinner landing on C is 1/4.

💡Okay, so I have this problem here where α and β are roots of the quadratic equation x² + 2017x + 1 = 0. I need to find the value of the expression (1 + 2020α + α²)(1 + 2020β + β²). The options given are A: 4, B: 9, C: 12, D: 15. Hmm, let me think about how to approach this.First, since α and β are roots of the equation x² + 2017x + 1 = 0, I remember that for any quadratic equation ax² + bx + c = 0, the sum of the roots is -b/a and the product is c/a. So in this case, the sum of the roots α + β should be -2017/1, which is -2017, and the product αβ should be 1/1, which is 1. Okay, so I have:α + β = -2017 αβ = 1Now, the expression I need to evaluate is (1 + 2020α + α²)(1 + 2020β + β²). That looks a bit complicated, but maybe I can simplify it somehow. Let me first look at each part inside the parentheses separately.Looking at 1 + 2020α + α², I notice that it's similar to the quadratic equation given, which is α² + 2017α + 1 = 0. Wait, that's interesting. If I rearrange the quadratic equation, I get α² = -2017α - 1. Maybe I can substitute that into the expression to simplify it.So, substituting α² = -2017α - 1 into 1 + 2020α + α², I get:1 + 2020α + (-2017α - 1) = 1 + 2020α - 2017α - 1 = (1 - 1) + (2020α - 2017α) = 0 + 3α = 3αOh, that's nice! So 1 + 2020α + α² simplifies to 3α. Similarly, I can do the same substitution for the other part of the expression.Looking at 1 + 2020β + β², since β is also a root of the same quadratic equation, I can substitute β² = -2017β - 1. So:1 + 2020β + β² = 1 + 2020β + (-2017β - 1) = 1 + 2020β - 2017β - 1 = (1 - 1) + (2020β - 2017β) = 0 + 3β = 3βSo now, the original expression simplifies to:(3α)(3β) = 9αβHmm, and I already know from earlier that αβ = 1. So substituting that in:9αβ = 9 * 1 = 9So the value of the expression is 9, which corresponds to option B.Wait, let me double-check my steps to make sure I didn't make any mistakes. Starting from the quadratic equation, I correctly identified the sum and product of the roots. Then, I recognized that α² and β² can be expressed in terms of α and β using the original equation. Substituting those into the given expression simplified each part to 3α and 3β, respectively. Multiplying them together gave me 9αβ, and since αβ is 1, the result is 9. Yeah, that seems correct.I don't see any errors in my reasoning, so I think the answer is indeed 9.**Final Answer**boxed{B}

💡First, I need to simplify the equation (frac{6x}{18} = frac{3}{x}). I can reduce (frac{6x}{18}) to (frac{x}{3}).Next, I'll eliminate the fractions by multiplying both sides of the equation by (3x). This gives me (x times 3x = 3x^2) on the left side and (3x times frac{3}{x} = 9) on the right side.Now, I have the equation (3x^2 = 9). To solve for (x), I'll divide both sides by 3, resulting in (x^2 = 3).Taking the square root of both sides, I find that (x = pm sqrt{3}).Finally, to find the sum of all solutions, I'll add (sqrt{3}) and (-sqrt{3}), which equals 0.

💡Okay, so I have this problem where there are 2008 students, and we need to select 50 of them to form a delegation. The process involves two steps: first, excluding 8 students using simple random sampling, and then selecting the remaining 2000 students using systematic sampling. The question is about the probability of each student being selected, and whether it's equal or not.Alright, let me break this down. First, simple random sampling (SRS) is used to exclude 8 students. That means every student has an equal chance of being excluded, right? So, the probability that any particular student is excluded is 8 out of 2008. That simplifies to 8/2008, which can be reduced further. Let me do that: 8 divided by 8 is 1, and 2008 divided by 8 is 251. So, it's 1/251. So, each student has a 1/251 chance of being excluded.Now, after excluding 8 students, we're left with 2000 students. From these 2000, we need to select 50 using systematic sampling. Hmm, systematic sampling is a method where you select every k-th individual from a list. The value of k is usually determined by dividing the population size by the sample size. So, in this case, k would be 2000 divided by 50, which is 40. So, every 40th student would be selected.But wait, how does systematic sampling affect the probability of each student being selected? In systematic sampling, once the starting point is randomly chosen, every k-th student is selected. So, each student has an equal chance of being selected because the starting point is random. That means the probability of being selected in the systematic sampling phase is 50/2000, which simplifies to 1/40.But hold on, the overall probability of a student being selected isn't just the probability of being selected in the second phase. It's also affected by not being excluded in the first phase. So, the overall probability should be the probability of not being excluded multiplied by the probability of being selected in the second phase.So, the probability of not being excluded in the first phase is 1 minus the probability of being excluded. We already found that the probability of being excluded is 1/251, so the probability of not being excluded is 1 - 1/251, which is 250/251.Then, the probability of being selected in the second phase is 1/40, as we calculated earlier. So, the overall probability is (250/251) * (1/40). Let me compute that:250 divided by 251 is approximately 0.996, and 1 divided by 40 is 0.025. Multiplying these together gives approximately 0.0249, which is roughly 1/40. But let me do it more precisely.250/251 * 1/40 = (250 * 1) / (251 * 40) = 250 / 10040. Simplifying this fraction, both numerator and denominator can be divided by 2: 125 / 5020. Dividing numerator and denominator by 5: 25 / 1004. So, the exact probability is 25/1004.Wait, so that's the same as 50/2008, which is the overall probability without considering the two-step process. Interesting. So, does that mean that each student has an equal probability of being selected, regardless of the two-step process?But hold on, in systematic sampling, isn't there a possibility that certain students have a higher or lower chance of being selected depending on the starting point? For example, if the starting point is randomly selected, then every student has an equal chance of being selected, right? Because the starting point is random, so the selection is uniform across the population.But in this case, we first exclude 8 students randomly, and then apply systematic sampling on the remaining 2000. So, does the exclusion affect the systematic sampling? Or does the exclusion just reduce the population size, and the systematic sampling is applied on the remaining?I think it does reduce the population size, so the systematic sampling is applied on 2000 students, and each of those 2000 has an equal chance of being selected, which is 50/2000 = 1/40. But then, the exclusion step affects the overall probability because not all students have the same chance of making it to the second phase.Wait, but the exclusion is done via simple random sampling, so every student has an equal chance of being excluded. So, the probability of being excluded is the same for all students, and the probability of not being excluded is also the same for all students. Therefore, the overall probability of being selected should be the same for all students.So, combining both steps, the probability of a student being selected is (probability of not being excluded) * (probability of being selected in systematic sampling). As we calculated earlier, that's (250/251) * (1/40) = 25/1004.Therefore, each student has an equal probability of being selected, and that probability is 25/1004.But wait, let me double-check. If we didn't exclude anyone, and just did systematic sampling on 2008 students to select 50, the probability would be 50/2008, which is the same as 25/1004. So, by excluding 8 students first and then doing systematic sampling, we're effectively doing the same as if we had just done systematic sampling on the original population. Because the exclusion is random, it doesn't introduce any bias.Therefore, the probability is equal for all students, and it's 25/1004.So, looking back at the options:A: Not all equalB: Equal to 25/1004C: All unequalD: Equal to 1/40So, the correct answer should be B: Equal to 25/1004.But wait, let me think again. In the systematic sampling, the probability is 1/40 for the 2000 students. But the overall probability is 25/1004, which is approximately 0.0249, which is slightly less than 1/40 (which is 0.025). So, is it exactly 25/1004 or approximately 1/40?But 25/1004 is exactly equal to 50/2008, which is the overall probability without considering the two-step process. So, since the exclusion is random, the overall probability remains the same as if we had just done systematic sampling on the entire population.Therefore, the probability is exactly 25/1004, which is option B.I think that makes sense. So, the answer should be B.

💡To solve for ( x ) in the equation ( frac{2x - 4}{x + 1} = 0 ), follow these steps:1. **Set the numerator equal to zero**: A fraction equals zero when its numerator is zero (provided the denominator is not zero). Therefore, set ( 2x - 4 = 0 ).2. **Solve for ( x )**: [ 2x - 4 = 0 2x = 4 x = frac{4}{2} x = 2 ]3. **Check the denominator**: Ensure that the denominator ( x + 1 ) is not zero to avoid division by zero. Substitute ( x = 2 ) into the denominator: [ 2 + 1 = 3 neq 0 ] Since the denominator is not zero, ( x = 2 ) is a valid solution.Thus, the value of ( x ) that satisfies the equation is ( 2 ).

💡Okay, so I have this problem where there are three different numbers a, b, and c. The equations x² + a x + 1 = 0 and x² + b x + c = 0 have a common real root. Additionally, the equations x² + x + a = 0 and x² + c x + b = 0 also have a common real root. I need to find the sum a + b + c.Hmm, let's break this down. First, I know that if two quadratic equations have a common real root, then that root must satisfy both equations. So, let's denote the common root of the first pair of equations as α. That means:α² + a α + 1 = 0 ...(1)andα² + b α + c = 0 ...(2)If I subtract equation (2) from equation (1), I get:(α² + a α + 1) - (α² + b α + c) = 0Simplifying this:(a - b) α + (1 - c) = 0So, (a - b) α = c - 1Therefore, α = (c - 1)/(a - b) ...(3)Okay, so that's the value of α in terms of a, b, and c.Now, let's look at the second pair of equations: x² + x + a = 0 and x² + c x + b = 0. Let's denote their common real root as β. So,β² + β + a = 0 ...(4)andβ² + c β + b = 0 ...(5)Subtracting equation (5) from equation (4):(β² + β + a) - (β² + c β + b) = 0Simplifying:(1 - c) β + (a - b) = 0So, (1 - c) β = b - aTherefore, β = (b - a)/(1 - c) ...(6)Now, interestingly, from equation (3) and (6), we can see that α and β are related. Let's see:From equation (3): α = (c - 1)/(a - b)From equation (6): β = (b - a)/(1 - c) = (a - b)/(c - 1)Wait, that's interesting. So, β = (a - b)/(c - 1) = 1/αBecause α = (c - 1)/(a - b), so 1/α = (a - b)/(c - 1) = βSo, β = 1/αThat's a useful relationship. So, the common roots α and β are reciprocals of each other.Now, let's think about the first equation: x² + a x + 1 = 0. The product of its roots is 1 (since the constant term is 1). So, if one root is α, the other root must be 1/α.Similarly, for the equation x² + x + a = 0, the product of its roots is a. If one root is β = 1/α, then the other root must be a/(1/α) = a α.But wait, let's check that.Wait, for the equation x² + x + a = 0, the product of the roots is a, and the sum of the roots is -1 (since the coefficient of x is 1, and the sum is -1).So, if one root is β = 1/α, then the other root is -1 - β = -1 - 1/α.But also, the product of the roots is a, so:β * (other root) = aSo, (1/α) * (-1 - 1/α) = aLet me compute that:(1/α) * (-1 - 1/α) = -1/α - 1/α² = aBut from equation (1), we have α² + a α + 1 = 0, so α² = -a α - 1So, 1/α² = 1/(-a α - 1)Let me substitute that into the expression for a:-1/α - 1/α² = aSo,-1/α - 1/(-a α - 1) = aLet me compute this:First term: -1/αSecond term: -1/(-a α - 1) = 1/(a α + 1)So,-1/α + 1/(a α + 1) = aLet me combine these terms:[ - (a α + 1) + α ] / [ α (a α + 1) ] = aSimplify numerator:- a α - 1 + α = (-a α + α) - 1 = α (1 - a) - 1So,[ α (1 - a) - 1 ] / [ α (a α + 1) ] = aMultiply both sides by denominator:α (1 - a) - 1 = a * α (a α + 1)Let me expand the right-hand side:a * α (a α + 1) = a² α² + a αSo, equation becomes:α (1 - a) - 1 = a² α² + a αBring all terms to one side:a² α² + a α - α (1 - a) + 1 = 0Simplify:a² α² + a α - α + a α + 1 = 0Combine like terms:a² α² + (a α + a α) - α + 1 = 0So,a² α² + 2 a α - α + 1 = 0Factor α from the middle terms:a² α² + α (2 a - 1) + 1 = 0Hmm, this seems complicated. Maybe there's another approach.Wait, let's recall that α is a root of x² + a x + 1 = 0, so α² = -a α - 1.Let me substitute α² = -a α - 1 into the equation we just derived:a² α² + 2 a α - α + 1 = 0Replace α²:a² (-a α - 1) + 2 a α - α + 1 = 0Expand:- a³ α - a² + 2 a α - α + 1 = 0Combine like terms:(-a³ α + 2 a α - α) + (-a² + 1) = 0Factor α from the first group:α (-a³ + 2 a - 1) + (-a² + 1) = 0So,α (-a³ + 2 a - 1) = a² - 1Thus,α = (a² - 1)/(-a³ + 2 a - 1)Simplify denominator:- a³ + 2 a - 1 = - (a³ - 2 a + 1)Let me factor the denominator:a³ - 2 a + 1. Let's try to factor this.Try a = 1: 1 - 2 + 1 = 0. So, (a - 1) is a factor.Divide a³ - 2 a + 1 by (a - 1):Using synthetic division:1 | 1 0 -2 1Bring down 1.Multiply by 1: 1Add to next term: 0 + 1 = 1Multiply by 1: 1Add to next term: -2 + 1 = -1Multiply by 1: -1Add to last term: 1 + (-1) = 0So, quotient is a² + a - 1Thus, a³ - 2 a + 1 = (a - 1)(a² + a - 1)So, denominator is - (a - 1)(a² + a - 1)Thus,α = (a² - 1)/[- (a - 1)(a² + a - 1)] = -(a² - 1)/[(a - 1)(a² + a - 1)]Note that a² - 1 = (a - 1)(a + 1), so:α = - (a - 1)(a + 1)/[(a - 1)(a² + a - 1)] = - (a + 1)/(a² + a - 1)So, α = - (a + 1)/(a² + a - 1)But from equation (3), we have α = (c - 1)/(a - b)So,(c - 1)/(a - b) = - (a + 1)/(a² + a - 1)Let me write this as:(c - 1)/(a - b) = - (a + 1)/(a² + a - 1)So,(c - 1) = - (a + 1)(a - b)/(a² + a - 1)Hmm, this is getting a bit messy. Maybe I should look for another relationship.Wait, earlier we found that β = 1/α. So, β = 1/α.From equation (6): β = (b - a)/(1 - c) = (a - b)/(c - 1)But β is also a root of x² + x + a = 0, so:β² + β + a = 0But β = 1/α, so:(1/α)² + (1/α) + a = 0Multiply through by α²:1 + α + a α² = 0But from equation (1): α² = -a α - 1So,1 + α + a (-a α - 1) = 0Simplify:1 + α - a² α - a = 0Factor α:1 - a + α (1 - a²) = 0So,α (1 - a²) = a - 1Thus,α = (a - 1)/(1 - a²) = (a - 1)/[-(a² - 1)] = - (a - 1)/(a² - 1)Simplify:- (a - 1)/(a - 1)(a + 1) = -1/(a + 1)So, α = -1/(a + 1)Wait, that's a simpler expression for α.So, α = -1/(a + 1)But earlier, we had α = - (a + 1)/(a² + a - 1)So,-1/(a + 1) = - (a + 1)/(a² + a - 1)Multiply both sides by -1:1/(a + 1) = (a + 1)/(a² + a - 1)Cross-multiplying:a² + a - 1 = (a + 1)²Compute (a + 1)²:a² + 2 a + 1So,a² + a - 1 = a² + 2 a + 1Subtract a² from both sides:a - 1 = 2 a + 1Subtract a from both sides:-1 = a + 1Subtract 1:-2 = aSo, a = -2Okay, so we found that a = -2.Now, let's find α.From α = -1/(a + 1) = -1/(-2 + 1) = -1/(-1) = 1So, α = 1So, the common root α is 1.Now, let's find c and b.From equation (3): α = (c - 1)/(a - b)We have α = 1, a = -2So,1 = (c - 1)/(-2 - b)Multiply both sides by (-2 - b):-2 - b = c - 1So,c = -2 - b + 1 = -1 - bSo, c = -1 - bNow, let's use the fact that α = 1 is a root of x² + b x + c = 0.So,1² + b * 1 + c = 0So,1 + b + c = 0But c = -1 - b, so:1 + b + (-1 - b) = 0Simplify:1 + b - 1 - b = 00 = 0Hmm, that's an identity, so it doesn't give us new information.Let's look at the other pair of equations: x² + x + a = 0 and x² + c x + b = 0 have a common root β = 1/α = 1/1 = 1.Wait, β = 1 as well? Wait, no, β = 1/α = 1, but let's check.Wait, β is the common root of x² + x + a = 0 and x² + c x + b = 0.We found that β = 1/α = 1, so β = 1.So, let's substitute β = 1 into x² + x + a = 0:1² + 1 + a = 0So,1 + 1 + a = 0Thus,2 + a = 0So,a = -2Which is consistent with what we found earlier.Now, substitute β = 1 into x² + c x + b = 0:1² + c * 1 + b = 0So,1 + c + b = 0But we already have c = -1 - b, so:1 + (-1 - b) + b = 0Simplify:1 - 1 - b + b = 00 = 0Again, an identity.Hmm, seems like we need another equation to find b and c.Wait, let's recall that the equations x² + a x + 1 = 0 and x² + b x + c = 0 have a common root α = 1, so the other roots must satisfy the product.For x² + a x + 1 = 0, the product of roots is 1. Since one root is 1, the other root is 1/1 = 1. Wait, but that would mean both roots are 1, but the problem states that a, b, c are different numbers, so the quadratics must have distinct roots.Wait, hold on. If α = 1 is a root of x² + a x + 1 = 0, then the other root is 1 as well, because product is 1. But that would mean the quadratic has a double root at 1, which would imply that the quadratic is (x - 1)² = x² - 2x + 1. So, comparing with x² + a x + 1 = 0, we have a = -2, which is consistent.But then, the quadratic x² + b x + c = 0 also has root 1. Let's find the other root.The product of the roots is c, so if one root is 1, the other root is c.The sum of the roots is -b, so 1 + c = -bThus, c = -b - 1Which is consistent with what we found earlier.But we need another condition to find b and c.Wait, let's look at the other quadratic: x² + x + a = 0, which has root β = 1. So, substituting β = 1:1 + 1 + a = 0 => a = -2, which we already know.The other root of this quadratic is, let's call it γ, such that:Sum of roots: 1 + γ = -1 => γ = -2Product of roots: 1 * γ = a = -2 => γ = -2So, the other root is -2.Now, since the quadratic x² + c x + b = 0 has root β = 1, let's find its other root.Let the other root be δ.Sum of roots: 1 + δ = -cProduct of roots: 1 * δ = b => δ = bSo, sum of roots: 1 + b = -c => c = -1 - bWhich is again consistent.But we need another equation. Maybe the other root γ = -2 is related.Wait, is there any relationship between the roots of different quadratics?Wait, perhaps the other roots are related. Let's see.From x² + a x + 1 = 0, roots are 1 and 1 (but that would be a double root, but the problem states different numbers a, b, c, so maybe the quadratics have distinct roots? Wait, no, the problem says different numbers a, b, c, not necessarily the roots.Wait, but if a quadratic has a double root, that's still fine as long as a, b, c are different.But in our case, x² + a x + 1 = 0 has a double root at 1, so a = -2.Then, x² + b x + c = 0 has root 1 and another root c.Similarly, x² + x + a = 0 has roots 1 and -2.And x² + c x + b = 0 has roots 1 and b.But we need to ensure that all roots are real, which they are since we have common real roots.But I think we need to find another relationship.Wait, let's consider that the quadratic x² + c x + b = 0 has roots 1 and b.So, the quadratic can be written as (x - 1)(x - b) = x² - (1 + b)x + bBut the given quadratic is x² + c x + b = 0So, comparing coefficients:- (1 + b) = candb = bSo, c = - (1 + b)Which is consistent with what we had before.So, c = -1 - bSo, we have c = -1 - bBut we need another equation to find b and c.Wait, perhaps using the fact that the quadratic x² + b x + c = 0 has roots 1 and c.So, the quadratic can be written as (x - 1)(x - c) = x² - (1 + c)x + cBut the given quadratic is x² + b x + c = 0So, comparing coefficients:- (1 + c) = bandc = cSo, b = - (1 + c)But since c = -1 - b, substitute into this:b = - (1 + (-1 - b)) = - (1 -1 - b) = - (-b) = bSo, b = b, which is an identity.Hmm, so we're stuck again.Wait, maybe we need to consider the discriminant to ensure real roots.For x² + b x + c = 0, discriminant D = b² - 4c ≥ 0Similarly, for x² + c x + b = 0, discriminant D = c² - 4b ≥ 0So, we have:b² - 4c ≥ 0 ...(7)andc² - 4b ≥ 0 ...(8)But since c = -1 - b, substitute into these inequalities.First, inequality (7):b² - 4c = b² - 4(-1 - b) = b² + 4 + 4b ≥ 0So,b² + 4b + 4 ≥ 0Which is (b + 2)² ≥ 0, which is always true.Inequality (8):c² - 4b = (-1 - b)² - 4b = (1 + 2b + b²) - 4b = b² - 2b + 1 ≥ 0Which is (b - 1)² ≥ 0, which is also always true.So, no new information from discriminants.Hmm, seems like we need another approach.Wait, let's recall that the quadratic x² + x + a = 0 has roots 1 and -2, as we found earlier.So, the quadratic is (x - 1)(x + 2) = x² + x - 2Thus, a = -2, which is consistent.Now, the quadratic x² + c x + b = 0 has roots 1 and b.So, it can be written as (x - 1)(x - b) = x² - (1 + b)x + bBut the quadratic is x² + c x + b = 0, so:- (1 + b) = cSo, c = -1 - bWe already have this.So, we need another equation.Wait, perhaps the quadratic x² + b x + c = 0 has roots 1 and c, so it can be written as (x - 1)(x - c) = x² - (1 + c)x + cBut the quadratic is x² + b x + c = 0, so:- (1 + c) = bThus, b = -1 - cBut since c = -1 - b, substitute:b = -1 - (-1 - b) = -1 + 1 + b = bAgain, identity.Hmm, seems like we're going in circles.Wait, maybe we can assign a value to b and find c accordingly.But since a, b, c are different numbers, and a = -2, so b ≠ -2, c ≠ -2, and b ≠ c.Let me assume b = k, then c = -1 - k.We need to ensure that the quadratics have real roots, which they do as we've checked.But we need another condition to find k.Wait, perhaps the other roots of the quadratics must be real and different.Wait, but we already have that.Alternatively, maybe the other roots are related.Wait, the quadratic x² + b x + c = 0 has roots 1 and c.Similarly, the quadratic x² + c x + b = 0 has roots 1 and b.So, c is a root of x² + b x + c = 0, and b is a root of x² + c x + b = 0.So, substituting c into x² + b x + c = 0:c² + b c + c = 0Similarly, substituting b into x² + c x + b = 0:b² + c b + b = 0So, we have two equations:c² + b c + c = 0 ...(9)andb² + c b + b = 0 ...(10)Let me write them:From (9):c² + b c + c = 0 => c(c + b + 1) = 0From (10):b² + c b + b = 0 => b(b + c + 1) = 0So, from (9): c(c + b + 1) = 0From (10): b(b + c + 1) = 0So, either c = 0 or c + b + 1 = 0Similarly, either b = 0 or b + c + 1 = 0But since a = -2, and a, b, c are different, b ≠ -2, c ≠ -2.Case 1: c = 0Then from c = -1 - b, 0 = -1 - b => b = -1So, b = -1, c = 0Check if they are different from a = -2: yes, -1 ≠ -2, 0 ≠ -2Also, b ≠ c: -1 ≠ 0So, this is a valid case.Case 2: c + b + 1 = 0From c + b + 1 = 0, and c = -1 - b, substitute:(-1 - b) + b + 1 = 0 => (-1 - b + b + 1) = 0 => 0 = 0So, this is always true.Thus, in addition to case 1, we have case 2 where c + b + 1 = 0, which is always satisfied given c = -1 - b.But in case 1, c = 0 and b = -1, which also satisfies c + b + 1 = 0 + (-1) + 1 = 0So, essentially, case 1 is a specific instance of case 2.Therefore, the only solution is c = -1 - b, with no additional constraints.But we need to find specific values for b and c.Wait, but from the quadratics, we can find more.Wait, the quadratic x² + b x + c = 0 has roots 1 and c.So, sum of roots: 1 + c = -bProduct of roots: 1 * c = cWhich is consistent.Similarly, quadratic x² + c x + b = 0 has roots 1 and b.Sum of roots: 1 + b = -cProduct of roots: 1 * b = bConsistent.But we need another condition.Wait, perhaps the other roots of the quadratics must be real and distinct.But we already have that.Alternatively, maybe the quadratics x² + b x + c = 0 and x² + c x + b = 0 must have real roots, which they do as we've checked.Wait, but perhaps the other roots must not be equal to 1 or something.Wait, in the quadratic x² + b x + c = 0, the roots are 1 and c.Similarly, in x² + c x + b = 0, the roots are 1 and b.So, unless c = b, which is not allowed since a, b, c are different, and c = -1 - b, so c ≠ b unless -1 - b = b => -1 = 2b => b = -0.5, but then c = -1 - (-0.5) = -0.5, which would make c = b = -0.5, but a = -2, so they are different. Wait, but if b = -0.5, c = -0.5, which would make b = c, which is not allowed since a, b, c are different.Thus, b ≠ c, so c ≠ -1 - b only if b ≠ -0.5, but in our case, c = -1 - b, so if b = -0.5, c = -0.5, which is not allowed, so b ≠ -0.5.But this doesn't help us find specific values.Wait, maybe we can use the fact that the quadratic x² + b x + c = 0 has roots 1 and c, and the quadratic x² + c x + b = 0 has roots 1 and b.So, let's write both quadratics:x² + b x + c = (x - 1)(x - c) = x² - (1 + c)x + cThus, b = - (1 + c)Similarly,x² + c x + b = (x - 1)(x - b) = x² - (1 + b)x + bThus, c = - (1 + b)So, we have:b = - (1 + c)c = - (1 + b)Substitute c from the second equation into the first:b = - (1 + (-1 - b)) = - (1 -1 - b) = - (-b) = bSo, again, identity.Thus, we can't find unique values for b and c, except that c = -1 - b.But since a, b, c are different, and a = -2, we can choose any b ≠ -2, c = -1 - b ≠ -2, and b ≠ c.So, for example, if we choose b = 0, then c = -1 - 0 = -1Check if they are different: a = -2, b = 0, c = -1: all different.Alternatively, b = 1, c = -2, but c = -2 is equal to a, which is not allowed.So, b = 1 is invalid.Similarly, b = -1, c = 0: valid.So, possible solutions are:a = -2, b = 0, c = -1ora = -2, b = -1, c = 0But let's check if these satisfy all conditions.First, a = -2, b = 0, c = -1Check quadratics:x² + a x + 1 = x² - 2x + 1 = (x - 1)², roots 1 (double root)x² + b x + c = x² + 0x -1 = x² -1, roots 1 and -1So, common root 1: good.Now, x² + x + a = x² + x - 2, roots 1 and -2x² + c x + b = x² -1 x + 0 = x² - x, roots 0 and 1Common root 1: good.So, this works.Similarly, a = -2, b = -1, c = 0Check quadratics:x² + a x + 1 = x² - 2x + 1 = (x -1)², roots 1x² + b x + c = x² - x + 0 = x² -x, roots 0 and 1Common root 1: good.x² + x + a = x² + x -2, roots 1 and -2x² + c x + b = x² + 0x -1 = x² -1, roots 1 and -1Common root 1: good.So, both sets work.But in both cases, a + b + c = -2 + 0 + (-1) = -3 or -2 + (-1) + 0 = -3So, in both cases, the sum is -3.Therefore, the sum a + b + c is -3.