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(A) Given $\frac{x_1^5+x_2^5}{x_1^3+x_2^3} = \frac{11}{3}$.Using the identity $a^5+b^5 = (a^2+b^2)(a^3+b^3) - a^2b^2(a+b)$,we have:$\frac{(x_1^2+x_2^2

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$x^2 \pm 3x + 2 = 0$

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(A) Given $\frac{x_1^5+x_2^5}{x_1^3+x_2^3} = \frac{11}{3}$.Using the identity $a^5+b^5 = (a^2+b^2)(a^3+b^3) - a^2b^2(a+b)$,we have:$\frac{(x_1^2+x_2^2)(x_1^3+x_2^3) - x_1^2x_2^2(x_1+x_2)}{x_1^3+x_2^3} = \frac{11}{3}$$(x_1^2+x_2^2) - \frac{x_1^2x_2^2(x_1+x_2)}{(x_1+x_2)(x_1^2-x_1x_2+x_2^2)} = \frac{11}{3}$Since $x_1^2+x_2^2 = 5$,we get $5 - \frac{x_1^2x_2^2}{5-x_1x_2} = \frac{11}{3}$.Let $x_1x_2 = t$. Then $5 - \frac{t^2}{5-t} = \frac{11}{3}$.$3(25 - 5t - t^2) = 55 - 11t \Rightarrow 3t^2 + 4t - 20 = 0$.Solving for $t$,we get $t = 2$ or $t = -10/3$.If $t = 2$,$(x_1+x_2)^2 = x_1^2+x_2^2 + 2x_1x_2 = 5 + 2(2) = 9$,so $x_1+x_2 = \pm 3$.If $t = -10/3$,$(x_1+x_2)^2 = 5 + 2(-10/3) = -5/3 Thus,the quadratic equation is $x^2 - (x_1+x_2)x + x_1x_2 = 0$,which gives $x^2 \pm 3x + 2 = 0$.

Which of the following quadratic equations has real roots $x_1, x_2$ that satisfy the conditions $x_1^2+x_2^2=5$ and $3(x_1^5+x_2^5)=11(x_1^3+x_2^3)$?

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