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(A) For a quadratic equation $ax^2 + bx + c = 0$,the product of the roots is given by $\frac{c}{a}$.Comparing the given equation $2x^2 + 6x + (\alpha^

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(A) For a quadratic equation $ax^2 + bx + c = 0$,the product of the roots is given by $\frac{c}{a}$.Comparing the given equation $2x^2 + 6x + (\alpha^2 + 1) = 0$ with the standard form,we have $a = 2$ and $c = \alpha^2 + 1$.The product of the roots is $\frac{\alpha^2 + 1}{2}$.Given that the product of the roots is $-\alpha$,we have $\frac{\alpha^2 + 1}{2} = -\alpha$.Multiplying both sides by $2$,we get $\alpha^2 + 1 = -2\alpha$.Rearranging the terms,we get $\alpha^2 + 2\alpha + 1 = 0$.This is a perfect square: $(\alpha + 1)^2 = 0$.Thus,$\alpha = -1$.

If the product of the roots of the equation $2x^2 + 6x + \alpha^2 + 1 = 0$ is $-\alpha$,then the value of $\alpha$ will be

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