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(D) Given that $\alpha$ and $\beta$ are the roots of the equation $x^2+bx+c=0$. From the relation between roots and coefficients,we have $\alpha+\beta

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$\frac{1}{2}(b-q)$

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(D) Given that $\alpha$ and $\beta$ are the roots of the equation $x^2+bx+c=0$. From the relation between roots and coefficients,we have $\alpha+\beta = -b$. Also,$\alpha+h$ and $\beta+h$ are the roots of the equation $x^2+qx+r=0$. Therefore,the sum of the roots is $(\alpha+h) + (\beta+h) = -q$. This simplifies to $(\alpha+\beta) + 2h = -q$. Substituting $\alpha+\beta = -b$,we get $-b + 2h = -q$. Rearranging the terms,$2h = b-q$. Thus,$h = \frac{1}{2}(b-q)$.

If $\alpha, \beta$ are the roots of the equation $x^2+bx+c=0$ and $\alpha+h, \beta+h$ are the roots of the equation $x^2+qx+r=0$,then $h$ is equal to

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