2
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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?

$\endgroup$
1
$\begingroup$

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

$\endgroup$
1
$\begingroup$

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

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 $a$?

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.