Given that equation is a positive constant, equal roots, find value of k

Question

I am having trouble solving this equation. It reads...

Given that the equation $kx^2+12x+k = 0$, where $k$ is a positive constant, has equal roots, find the value of $k$.

I am not sure where to start as there is two unknown variables, $k$ and $x$ so I don't think I can factorise. Any help would be much appreciated.

Answer 1

Hint: as $k>0$ is fixed, this equation is a second degree equation in the variable $x$. There is one root iff the discriminant is 0. In you case: $$ 12^2- 4\times k\times k = 0 $$

Answer 2

$p(x)=kx^2+12x+k=\text { (double root) }k(x-a)^2=kx^2-2kax+ka^2$

Equate coefficients so that $a^2=1$ and $2ka=-12$

Note that the general form of a quadratic with a double root is $a(x-b)^2=0$ with a double root at $x=b$.