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(D) Let the zeroes of the quadratic polynomial $p(x) = x^{2} + ax + b$ be $\alpha$ and $-\alpha$.
Sum of zeroes = $-\frac{	ext{coefficient of } x}{\

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has no linear term and the constant term is negative

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(D) Let the zeroes of the quadratic polynomial $p(x) = x^{2} + ax + b$ be $\alpha$ and $-\alpha$.
Sum of zeroes = $-\frac{	ext{coefficient of } x}{	ext{coefficient of } x^{2}} = -a$.
Since $\alpha + (-\alpha) = 0$,we have $0 = -a$,which implies $a = 0$.
Product of zeroes = $\frac{	ext{constant term}}{	ext{coefficient of } x^{2}} = b$.
Since $\alpha 	imes (-\alpha) = -\alpha^{2}$,we have $-\alpha^{2} = b$.
Since $\alpha^{2}$ is always positive for non-zero $\alpha$,$- \alpha^{2}$ must be negative,so $b < 0$.
Therefore,the polynomial has no linear term $(a = 0)$ and the constant term is negative $(b < 0)$.

If one of the zeroes of a quadratic polynomial of the form $x^{2}+ax+b$ is the negative of the other,then it

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