For the quadratic equation 2x^2 - 2(p - 2)x - | vedclass

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(B) Let the roots of the quadratic equation $2x^2 - 2(p - 2)x - p - 1 = 0$ be $\alpha$ and $\beta$.From the properties of roots,we have $\alpha + \bet

Vedclass · Accepted Answer

$3/2$

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(B) Let the roots of the quadratic equation $2x^2 - 2(p - 2)x - p - 1 = 0$ be $\alpha$ and $\beta$.From the properties of roots,we have $\alpha + \beta = p - 2$ and $\alpha \beta = \frac{-(p + 1)}{2}$.Let $S$ be the sum of the squares of the roots: $S = \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha \beta$.Substituting the values,we get $S = (p - 2)^2 - 2 \left( \frac{-(p + 1)}{2} \right) = (p - 2)^2 + p + 1$.Expanding this,$S = p^2 - 4p + 4 + p + 1 = p^2 - 3p + 5$.To minimize $S$,we complete the square: $S = \left( p - \frac{3}{2} \right)^2 + 5 - \frac{9}{4} = \left( p - \frac{3}{2} \right)^2 + \frac{11}{4}$.The expression $S$ is minimized when $p - \frac{3}{2} = 0$,which gives $p = \frac{3}{2}$.

For the quadratic equation $2x^2 - 2(p - 2)x - p - 1 = 0$,what should be the value of $p$ such that the sum of the squares of its roots is minimized?

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