Question

If $5p^2-7p-3=0$ and $5q^2-7q-3=0,p\ne q$, then the equation whose roots are $5p-4q$ and $5q-4p$ is

• A. $5x^2+7x-439=0$
• B. $5x^2-7x-439=0$
• C. $5x^2+7x+439=0$
• D. $5x^2-7x+439=0$

Answer 1

The correct option is B $5x^2-7x-439=0$

Given
$5p^2-7p-3=\mathrm{0....}\left(i\right)$
$5q^2-7q-3=\mathrm{0.....}\left(ii\right)$
It is clear from Eqs(i) and (ii) $p$ and $q$ satisfy the equation $5x^2-7x-3=0$
$\therefore p+q=\frac{7}{5};pq=\frac{-3}{5}...\left(iii\right)$
We have to find the equation whose roots are
$(5p-4q)$ and $(5q-4p)$
Here, sum of roots $=(5p-4q)+(5q-4p)=p+q=\frac{7}{5}$
and product of roots
$=(5p-4q)(5q-4p)=81pq-20{(p+q)}^2=\frac{-439}{5}$
$\therefore$ Required equation will be
$x^2-\left(sumofroots\right)x+productofroots=0$
$\Rightarrow 5x^2-7x-439=0$