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 find the value of $p+q$.

Given: Roots of equation $x^2+x+1=0$ are $a,\ b$ and roots of $x^2+px+q=0$ are $\frac{a}{b},\ \frac{b}{a}$

To do: To find the value of $p+q$.

Solution:

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

$\therefore$ Sum of roots, $a+b=-\frac{1}{1}=-1$

Product of roots, $ab=\frac{1}{1}=1\ .........\ ( i)$

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

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

Product of roots, $\frac{a}{b}\times\frac{b}{a}=q$

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

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

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

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

$\Rightarrow \frac{( -1)^2-2( 1)}{1}=-p$    [from equation (i)]

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

$\Rightarrow p=1$

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

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Updated on: 10-Oct-2022

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