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Roots of a Quadratic Equation

If α,β are ...

Question

If $α, β$ are the roots of the equation $a x^{2} + b x + c = 0$ , then form an equation whose roots are:
$(α - β)^{2}, (α + β)^{2}$

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Solution

Since $α$ and $β$ are the roots of the equation $a x^{2} + b x + c = 0$ , then,

$α + β = - \frac{b}{a}$

And,

$α β = \frac{c}{a}$

If the roots of any equation is ${(α - β)}^{2}$ and ${(α + β)}^{2}$ , then,

$(x - {(α - β)}^{2}) (x - {(α + β)}^{2}) = 0$

$(x - (α^{2} + β^{2} - 2 α β)) (x - (α^{2} + β^{2} + 2 α β)) = 0$

$(x - α^{2} - β^{2} + 2 α β) (x - α^{2} - β^{2} - 2 α β) = 0$

$((x - α^{2} - β^{2}) + 2 α β) ((x - α^{2} - β^{2}) - 2 α β) = 0$

$({(x - α^{2} - β^{2})}^{2} - {(2 α β)}^{2}) = 0$

$x^{2} + α^{2} + β^{2} - 2 α x + 2 α β - 2 x β - 4 {(α β)}^{2} = 0$

$x^{2} + (α^{2} + β^{2} + 2 α β) - 2 x (α + β) - 4 {(α β)}^{2} = 0$

$x^{2} + {(α + β)}^{2} - 2 x (α + β) - 4 {(α β)}^{2} = 0$

$x^{2} + {(- \frac{b}{a})}^{2} - 2 x (- \frac{b}{a}) - 4 {(\frac{c}{a})}^{2} = 0$

$x^{2} + \frac{b^{2}}{a^{2}} + 2 \frac{b}{a} x - 4 \frac{c^{2}}{a^{2}} = 0$

$a^{2} x^{2} + 2 a b x - 4 c^{2} + b^{2} = 0$

Therefore, the required equation is $a^{2} x^{2} + 2 a b x - 4 c^{2} + b^{2} = 0$ .

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