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

Answer 1

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

$5p^2-7p-3=0$
and $5q^2-7q-3=0$
So $p,q$ are the roots of the equation of
$5x^2-7x-3=0$
$\Rightarrow p+q=\frac{7}{5},pq=\frac{-3}{5}$

Let $a=5p-4q,b=5q-4p$ be the roots of a new quadratic equation.
So, sum of roots, $a+b=5(p+q)-4(p+q)\Rightarrow a+b=p+q=\frac{7}{5}$

Product of roots, $ab=(5p-4q)(5q-4p)$
$\Rightarrow ab=41pq-20(p^2+q^2)\Rightarrow ab=41pq-20(p+q{)}^2+40pq\Rightarrow ab=81pq-20(p+q{)}^2=\frac{-439}{5}$

Hence, the required equation is
$5x^2-7x-439=0$