Question

Can the quadratic polynomial${\mathrm{x}}^2+\mathrm{kx}+\mathrm{k}$have equal zeroes for some odd integer $\mathrm{k}>1$?

Answer 1

A quadratic equation will have equal roots if it satisfies the condition given below:

$\mathrm{b}²–4\mathrm{ac}=0$

Given equation is

$$\begin{gathered} \mathrm{x}²+\mathrm{kx}+\mathrm{k}=0 \\ \mathrm{a}=1,\mathrm{b}=\mathrm{k},\mathrm{x}=\mathrm{k} \end{gathered}$$

On substituting in the equation we get,

$$\begin{gathered} \mathrm{k}²–4(1)(\mathrm{k})=0 \\ \mathrm{k}²–4\mathrm{k}=0 \\ \mathrm{k}(\mathrm{k}–4)=0 \\ \mathrm{k}=0,\mathrm{k}=4 \end{gathered}$$

But by the given data we get to know that $\mathrm{k}$ is greater than $1$.

$\Rightarrow$ The value of $\mathrm{k}$ is $4$ if the equation has common roots.

Hence, if the value of$\mathrm{k}=4$, then the equation$(\mathrm{x}²+\mathrm{kx}+\mathrm{k})$will have equal roots.