If one of the roots of the equation x^2+px+q= | vedclass

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(D) Given equation is $x^2+px+q=0$ ... $(i)$Comparing with $ax^2+bx+c=0$,we get $a=1, b=p, c=q$.Let the roots be $\alpha$ and $\beta$. Given $\alpha =

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$p(3q-p^2)=q(q+1)$

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(D) Given equation is $x^2+px+q=0$ ... $(i)$Comparing with $ax^2+bx+c=0$,we get $a=1, b=p, c=q$.Let the roots be $\alpha$ and $\beta$. Given $\alpha = \beta^2$.From the relation between roots and coefficients:Sum of roots: $\alpha + \beta = -p \Rightarrow \beta^2 + \beta = -p$ ... (ii)Product of roots: $\alpha \beta = q$ $\Rightarrow \beta^2 \cdot \beta = q$ $\Rightarrow \beta^3 = q$ ... (iii)Cubing equation (ii) on both sides:$(\beta^2 + \beta)^3 = (-p)^3$$\beta^6 + \beta^3 + 3\beta^2 \cdot \beta(\beta^2 + \beta) = -p^3$Substitute $\beta^3 = q$ and $\beta^2 + \beta = -p$:$(q)^2 + q + 3q(-p) = -p^3$$q^2 + q - 3pq = -p^3$$p^3 - 3pq = -q^2 - q$$p(p^2 - 3q) = -q(q+1)$$p(3q - p^2) = q(q+1)$

If one of the roots of the equation $x^2+px+q=0$ is equal to the square of the other,then:

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