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The zeroes of the quadratic polynomial \[{{x}^{2}}+kx+k,\] \[k\ne 0\]:                  (NCERT EXEMPLAR)

A) cannot both be positive

B) cannot both be negative

C) are always unequal

D) are always equal

Correct Answer: A

Solution :

[a] Let \[p(x)={{x}^{2}}+kx+k,\] \[k\ne 0\]

On comparing \[p(x)\] with \[a{{x}^{2}}+bx+c,\] we get

\[a=1,\,\,b=k\]and \[c=k\]

Now,   \[x=\frac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\]

[by quadratic formula]

\[=\frac{-k\pm \sqrt{{{k}^{2}}-4k}}{2\times 1}\]

\[=\frac{-k\pm \sqrt{k(k-4)}}{2},\,\,\,k\ne 0\]

Here, we see that \[k(k-4)>0\]

\[\Rightarrow \,\,\,\,\,\,\,\,\,k\in (-\infty ,0)\cup (4,\infty )\]

Case l: lf \[k\in (-\infty ,0)\]i.e., \[k<0\]

\[\Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,a=1>0;\,b,c=k<0\]

So, both zeroes are of opposite sign.

Case II: If  \[k\in (4,\infty )\] i.e., \[k>4\]

\[\Rightarrow \,\,\,\,\,\,\,\,\,\,a=1>0;\,\,\,b,\,\,c=k>4\]

So, both zeroes are negative.

Hence, in any case zeroes of the given quadratic polynomial cannot be both positive.