Which of the following quadratic equations whose real roots $x_1, x_2$ satisfy the conditions $x_1^2+x_2^2=5 ; 3\left(x_1^5+x_2^5\right)=11\left(x_1^3+x_2^3\right) ?$

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Aicharya · Accepted Answer

$\frac{x_1^5+x_2^5}{x_1^3+x_2^3}=\frac{11}{3}$ $\begin{aligned} \Rightarrow & \frac{\left(x_1^2+x_2^2ight)\left(x_1^3+x_2^3ight)-x_1^2 x_2^2\left(x_1+x_2ight)}{x_1^3+x_2^3}=\frac{11}{3} \ & {\left[\because a^5+b^5=\left(a^2+b^2ight)\left(a^3+b^3ight)-a^2 b^2(a+b)ight] }\end{aligned}$ $\Rightarrow \quad\left(x_1^2+x_2^2ight)-\frac{x_1^2 x_2^2\left(x_1+x_2ight)}{\left(x_1+x_2ight)\left(x_1^2-x_1 x_2+x_2^2ight)}=\frac{11}{3}$ $\Rightarrow \quad 5-\frac{x_1^2 x_2^2}{5-x_1 x_2}=\frac{11}{3}$ Let $x_1 x_2=t$ $\begin{array}{ll}\Rightarrow & 5-\frac{t^2}{5-t}=\frac{11}{3} \ \Rightarrow & 3\left(25-5 t-t^2ight)=55-11 t\end{array}$ $\begin{aligned} \Rightarrow & & 3 t^2+4 t-20 & =0 \ \Rightarrow & & t & =2,-\frac{10}{3} \ \Rightarrow & & x_1 x_2 & =2, \frac{-10}{3}\end{aligned}$ When $x_1 x_2=2$ $\left(x_1+x_2ight)^2=x_1^2+2 x_1 x_2+x_2^2$ $=5+2 	imes 2=9$ $\Rightarrow \quad x_1+x_2= \pm 3$ When $x_1 x_2=-\frac{10}{3}$ $\left(x_1+x_2ight)^2=5-\frac{-20}{3}=-\frac{5}{3} < 0$ which is not possible. $	herefore \quad x_1+x_2= \pm 3$ Required quadratic equation $x^2 \pm 3 x+2=0$

Which of the following quadratic equations whose real roots $x_1, x_2$ satisfy the conditions $x_1^2+x_2^2=5…

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