Question

If the roots of the quadratic equation $k{x}^2+(a+b)x+ab$ are $(-1,-b)$ the value of $k$ is ?

Answer 1

Compute the required value:

Given: Roots of $k{x}^2+(a+b)x+ab$ are $(-1,-b)$.

So, substitute $x=-b$ in the quadratic equation.

$$\begin{gathered} k{(-b)}^2+(a+b)(-b)+ab=0 \\ \Rightarrow k{b}^2-ab-b^2+ab=0 \\ \Rightarrow k{b}^2=b^2 \\ \Rightarrow k=1 \end{gathered}$$

Hence, the value of $k$ is $1$.