Question

Find a quadratic equation whose roots $x_1$ and $x_2$ satisfy the condition
$x_1^2+x_2^2=5,3(x_1^5+x_2^5)=11(x_1^3+x_2^3)$.
(Assume that $x_1,x_2$ are real)

• A. $x^2\pm 3x+2=0$
• B. $x^2\pm 6x+12=0$
• C. $x^2\pm 9x+12=0$
• D. None of these

Answer 1

Answer: (A)

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

We have $3(x_1^5+x_2^5)=11(x_1^3+x_2^3)$
$\Rightarrow \frac{x_1^5+x_2^5}{x_1^3+x_2^3}=\frac{11}{3}$
$\Rightarrow \frac{(x_1^2+x_2^2)(x_1^3+x_2^3)-x_1^2{x}_2^2(x_1+x_2)}{(x_1^3+x_2^3)}=\frac{11}{3}$
$\Rightarrow (x_1^2+x_2^2)-\frac{x_1^2{x}_2^2(x_1+x_2)}{(x_1+x_2)(x_1^2+x_2^2-x_1{x}_2)}=\frac{11}{3}$
$\Rightarrow 5-\frac{x_1^2{x}_2^2}{5-x_1{x}_2}=\frac{11}{3}$
$\Rightarrow \frac{4}{3}=\frac{x_1^2{x}_2^2}{5-x_1{x}_2}$
$\Rightarrow 3x_1^2{x}_2^2+4x_1{x}_2-20=0$$\Rightarrow x_1{x}_2=2$ and $x_1{x}_2=-\frac{10}{3}$
If $x_1{x}_2=2$, ${x_1}^2+{x_2}^2=5$ gives $x_1+x_2=\pm 3$
If $x_1{x}_2=2$,${x_1}^2+{x_2}^2=5$ gives $(x_1+x_2{)}^2<0$ which is not possible.

Therefore, the correct answer is A.