If the equations x^2 - cx + d = 0 and x^2 - a | vedclass

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(C) Let the roots of $x^2 - cx + d = 0$ be $\alpha$ and $\beta$. Let the roots of $x^2 - ax + b = 0$ be $\alpha$ and $\alpha$ (since it has two equal

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$ac$

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(C) Let the roots of $x^2 - cx + d = 0$ be $\alpha$ and $\beta$. Let the roots of $x^2 - ax + b = 0$ be $\alpha$ and $\alpha$ (since it has two equal roots). From the second equation,the sum of roots is $2\alpha = a$,so $\alpha = \frac{a}{2}$. The product of roots is $\alpha^2 = b$,so $(\frac{a}{2})^2 = b$,which means $b = \frac{a^2}{4}$. Since $\alpha$ is also a root of the first equation,$\alpha^2 - c\alpha + d = 0$. Substituting $\alpha = \frac{a}{2}$,we get $(\frac{a}{2})^2 - c(\frac{a}{2}) + d = 0$. This simplifies to $\frac{a^2}{4} - \frac{ac}{2} + d = 0$. Since $b = \frac{a^2}{4}$,we substitute $b$ into the equation: $b - \frac{ac}{2} + d = 0$. Therefore,$b + d = \frac{ac}{2}$,which implies $2(b + d) = ac$.

If the equations $x^2 - cx + d = 0$ and $x^2 - ax + b = 0$ have one common root and the second equation has two equal roots,then $2(b + d) = \dots$

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