Let p(x)=x2+kx+k,k≠0
On comparing p(x) with ax2+bx+c, we get
a = 1, b = 5 and c = 5
Now, x=−b±√b2−4ac2a
= −k±√k2−4ac2×1
= −5±√k(k−4)2,k≠0
Here, we see that
K(k–4)>0
⇒ K ϵ (−∞,0)∪(4,∞)
Now, we know that
In quadratic polynomial ax2+bx+c
If a>o,b>0,c>0 or a<0,b<0,c<0
Then the polynomial has always all negative zeroes
And if a>0,c<0 or a<0,c>0, then the polynomial has always zeroes of opposite sign
Case I if k ϵ (−∞,0)I,e.,k<0
⇒ a=1>0,b,c=k<0
So, both zeroes are of opposite sign
Case II if k ϵ (4,∞) i.e., k>4
⇒ a=1>0,b,c≥4
So, both zeroes are negative
Hence, in any case zeroes of the given quadratic polynomial cannot both be positive