If alpha and beta are roots of ax^{2}+bx+c=0, | vedclass

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Question

(A) Given that $\alpha$ and $\beta$ are roots of $ax^{2}+bx+c=0$.Sum of roots: $\alpha+\beta = -\frac{b}{a}$.Product of roots: $\alpha\beta = \frac{c}

Vedclass · Accepted Answer

$a^{2}x^{2}-(b^{2}-2ac)x+c^{2}=0$

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(A) Given that $\alpha$ and $\beta$ are roots of $ax^{2}+bx+c=0$.Sum of roots: $\alpha+\beta = -\frac{b}{a}$.Product of roots: $\alpha\beta = \frac{c}{a}$.For the new equation with roots $\alpha^{2}$ and $\beta^{2}$:Sum of roots: $\alpha^{2}+\beta^{2} = (\alpha+\beta)^{2}-2\alpha\beta = (-\frac{b}{a})^{2}-2(\frac{c}{a}) = \frac{b^{2}}{a^{2}}-\frac{2c}{a} = \frac{b^{2}-2ac}{a^{2}}$.Product of roots: $\alpha^{2}\beta^{2} = (\alpha\beta)^{2} = (\frac{c}{a})^{2} = \frac{c^{2}}{a^{2}}$.The required quadratic equation is $x^{2}-(	ext{sum of roots})x + (	ext{product of roots}) = 0$.$x^{2}-(\frac{b^{2}-2ac}{a^{2}})x + \frac{c^{2}}{a^{2}} = 0$.Multiplying by $a^{2}$, we get $a^{2}x^{2}-(b^{2}-2ac)x+c^{2}=0$.

If $\alpha$ and $\beta$ are roots of $ax^{2}+bx+c=0$, then the equation whose roots are $\alpha^{2}$ and $\beta^{2}$ is

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