Question

Question 8
The zeros of the quadratic polynomial $x^2+kx+kwherek\ne 0$
(A) cannot both be positive
(B) cannot both be negative
(C) are always unequal
(D) are always equal

Answer 1

Let $p\left(x\right)=x^2+kx+k,k\ne 0$
On comparing p(x) with $a{x}^2+bx+c$, we get
a = 1, b = 5 and c = 5
Now, $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$
$=\frac{-k\pm \sqrt{k^2-4ac}}{2\times 1}$
$=\frac{-5\pm \sqrt{k(k-4)}}{2},k\ne 0$
Here, we see that
$K(k–4)>0$
$\Rightarrow Kϵ(-\infty ,0)\cup (4,\infty )$
Now, we know that
In quadratic polynomial $a{x}^2+bx+c$
If $a>o,b>0,c>0ora<0,b<0,c<0$
Then the polynomial has always all negative zeroes
And if $a>0,c<0ora<0,c>0$, then the polynomial has always zeroes of opposite sign
Case I if $kϵ(-\infty ,0)I,e.,k<0$
$\Rightarrow a=1>0,b,c=k<0$
So, both zeroes are of opposite sign
Case II if $kϵ(4,\infty )i.e.,k>4$
$\Rightarrow a=1>0,b,c\ge 4$
So, both zeroes are negative
Hence, in any case zeroes of the given quadratic polynomial cannot both be positive