If alpha and beta are the roots of x^2+3(a+3) | vedclass

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(B) For the quadratic equation $x^2+3(a+3)x-9a=0$ to have equal roots,the discriminant $D$ must be $0$.$D = [3(a+3)]^2 - 4(1)(-9a) = 0$$9(a^2+6a+9) +

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$-\frac{69}{4}$

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(B) For the quadratic equation $x^2+3(a+3)x-9a=0$ to have equal roots,the discriminant $D$ must be $0$.$D = [3(a+3)]^2 - 4(1)(-9a) = 0$$9(a^2+6a+9) + 36a = 0$$9a^2 + 54a + 81 + 36a = 0$$9a^2 + 90a + 81 = 0$$a^2 + 10a + 9 = 0$$(a+9)(a+1) = 0 \Rightarrow a = -9, -1$.For $a = -9$,the equation is $x^2 - 18x + 81 = 0$ $\Rightarrow (x-9)^2 = 0$ $\Rightarrow x = 9$. Let $\alpha = 9$.For $a = -1$,the equation is $x^2 + 6x + 9 = 0$ $\Rightarrow (x+3)^2 = 0$ $\Rightarrow x = -3$. Let $\beta = -3$.We want the minimum value of the expression $f(x) = x^2 + \alpha x - \beta = x^2 + 9x - (-3) = x^2 + 9x + 3$.Completing the square: $f(x) = (x + \frac{9}{2})^2 + 3 - \frac{81}{4} = (x + \frac{9}{2})^2 - \frac{69}{4}$.Since $(x + \frac{9}{2})^2 \geq 0$,the minimum value is $-\frac{69}{4}$.

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