Question

Find the value(s) of k so that the quadratic polynomial kx^2+x+k has equal zeroes.

Answer 1

here is ur answer....
a=k
b=1
c=k
so, D=b²-4ac
=(1)² -4*k*k
= 1-4k²
so.D=0
1-4k²=0
-4k²=-1
k²= 1/4
k= √1/4=1/2
Therefore, K=1/2

Answer 2

Answer:

The value of k is either $\frac{1}{2}$ or $-\frac{1}{2}$.

Step-by-step explanation:

The given polynomial is

$p(x)=kx^2+x+k$

A polynomial $f(x)=ax^2+bx+c$ has equal roots if

$b^2-4ac=0$

$1^2-4(k)(k)=0$

$1-4k^2=0$

$1=4k^2$

$\frac{1}{4}=k^2$

$\pm\sqrt{\frac{1}{4}}=k$

$\pm\frac{1}{2}=k$

Therefore the value of k is either $\frac{1}{2}$ or $-\frac{1}{2}$.