If alpha and beta are the roots of {x^2} + px | vedclass

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(C) For the equation ${x^2} + px + q = 0$,the sum of roots is $\alpha + \beta = -p$ and the product of roots is $\alpha \beta = q$.For the equation ${

Vedclass · Accepted Answer

${p^2} - 4q = {r^2} - 4s$

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(C) For the equation ${x^2} + px + q = 0$,the sum of roots is $\alpha + \beta = -p$ and the product of roots is $\alpha \beta = q$.For the equation ${x^2} + rx + s = 0$,the roots are $\alpha + h$ and $\beta + h$. Thus,the sum of roots is $(\alpha + h) + (\beta + h) = -r$,which simplifies to $(\alpha + \beta) + 2h = -r$. Substituting $\alpha + \beta = -p$,we get $-p + 2h = -r$,so $h = \frac{p - r}{2}$.The product of roots for the second equation is $(\alpha + h)(\beta + h) = s$,which expands to $\alpha \beta + h(\alpha + \beta) + {h^2} = s$. Substituting the known values,we get $q + h(-p) + {h^2} = s$. Substituting $h = \frac{p - r}{2}$,we have $q - p\left( \frac{p - r}{2} \right) + \left( \frac{p - r}{2} \right)^2 = s$. Multiplying by $4$,we get $4q - 2p(p - r) + (p - r)^2 = 4s$,which simplifies to $4q - 2p^2 + 2pr + p^2 + r^2 - 2pr = 4s$. This results in $4q - p^2 + r^2 = 4s$,or ${p^2} - 4q = {r^2} - 4s$.

If $\alpha$ and $\beta$ are the roots of ${x^2} + px + q = 0$ and $\alpha + h$ and $\beta + h$ are the roots of ${x^2} + rx + s = 0$,then

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