Question

Question1 (v)
Answer the following and justify
v) Can the quadratic polynomial $x^2+kx+k$ have equal zeroes for some odd integer k > 1?

Answer 1

Let p(x) = $x^2+kx+k$
If p(x) has equal zeroes, then its discriminant $(b^2-4ac)$ should be zero
$\therefore$ $D=B^2–4AC=0$
On comparing p(x) with $A{x}^2+Bx+C$, we get
A = 1, B = k and C = k.
$\therefore$ On substituting the values we get:

$(k{)}^2–4(1\left)\right(k)=0$
$\Rightarrow k(k–4)=0$
$\Rightarrow k(k–4)=0$
$\Rightarrow k=0,4$
So, he quadratic polynomial p(x) will have equal zeroes only at k = 0, 4.

So, the quadratic polynomial $x^2+kx+k$ does not have equal zeroes for some odd integer k > 1.