Finding roots of a quadratic equation having same roots for all real values of a parameter

Question

The quadratic equation $(5a + 3k)x^2 + (2a-k)x+(a-2k)=0$, where a is a real constant and $k$ is a variable parameter, has the same roots for all real values of $k$. Find the two roots and the value of $a$.

My attempt: I have got the discriminant as $(-14a^2+25k^2+24ak)$, but could not proceed after that. Please help me to solve this question.

Answer 1

If $k=\frac a2$, then the quadratic becomes $\frac{13}2ax^2+\frac32ax=0$. It has two roots, one of which is $0$. But clearly $0$ is a root only when $k=\frac a2$. So, it's the other root which is a root for all such quadratics.

Can you take it from here?

Answer 2

Hint:

It is the equation of the quadratic which passes through $$5x^2+2x+1=0$$ and $$3x^2-x-2=0$$

So if the second equation is satisfied, the given equation will have same roots

Answer 3

$$(5a + 3k)x^2 + (2a-k)x+(a-2k)=0$$

Assume that $a \ne 0$.

When $k=0$, you get $5ax^2 + 2ax+ a=0$ and the roots are $x = \dfrac{-1 \pm 2i}{5}$.

When $k=\dfrac a2$, you get $ax(13x+3)=0$ and the roots are $x=0$ and $x = -\dfrac{3}{13}$.

You don't get the same roots if $a \ne 0$. So what must be the value of $a$?