Question

If roots of equation $x^2+x+1=0$ are $a$, $b$ and roots of $x^2+px+q=0$ are $\frac{a}{b}$, $\frac{b}{a}$; then the value of $p+q$ is

• A. $2$
• B. $\frac{\sqrt{2}+1}{2}$
• C. $-1$
• D. $1$

Answer 1

The correct option is A $2$

Given, roots of the equation $x^2+x+1=0$ are $a$ and $b$.

$\therefore$ Sum of roots, $a+b=\frac{-Coefficientofx}{Coefficientof{x}^2}=-1$

and product of roots, $ab=\frac{Constantterm}{Coefficientof{x}^2}=1$ ....(i)

Again, roots of the equation $x^2+px+q=0=\frac{a}{b}$ and $\frac{b}{a}$.

$\therefore$ Sum of roots, $\frac{a}{b}+\frac{b}{a}=-p$

and product of roots, $\frac{a}{b}\cdot \frac{b}{a}=q$

$\Rightarrow 1=q$ ...(i)

Now, $\frac{a}{b}+\frac{b}{a}=-p$

$\Rightarrow \frac{a^2+b^2}{ab}=-p$

$\Rightarrow \frac{{(a+b)}^2-2ab}{ab}=-p$ [from equation (i)]

$\Rightarrow \frac{{(-1)}^2-2\left(1\right)}{1}=-p$

$\Rightarrow \frac{1-2}{1}=-p$

$\Rightarrow p=1$

$\therefore p+q=1+1=2$.