The zeroes of the quadratic polynomial x^{2}+ | vedclass

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Question

(B) Let $p(x) = x^{2} + kx + k, k 
eq 0$.For the zeroes to be real,the discriminant $D$ must be greater than or equal to $0$.$D = b^{2} - 4ac = k^{2}

Vedclass · Accepted Answer

cannot both be positive

Vedclass · Answer

(B) Let $p(x) = x^{2} + kx + k, k 
eq 0$.For the zeroes to be real,the discriminant $D$ must be greater than or equal to $0$.$D = b^{2} - 4ac = k^{2} - 4k \geq 0$.$k(k - 4) \geq 0$.This implies $k \in (-\infty, 0] \cup [4, \infty)$. Since $k 
eq 0$,$k \in (-\infty, 0) \cup [4, \infty)$.Let the zeroes be $\alpha$ and $\beta$. Then:Sum of zeroes $\alpha + \beta = -k/1 = -k$.Product of zeroes $\alpha \beta = k/1 = k$.Case $I$: If $k \in (-\infty, 0)$,then $k Case $II$: If $k \in [4, \infty)$,then $k > 0$. Since the product $\alpha \beta = k > 0$,both zeroes have the same sign. The sum $\alpha + \beta = -k In both cases,the zeroes cannot both be positive.

The zeroes of the quadratic polynomial $x^{2}+kx+k, k \neq 0,$

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