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Nature of Roots

If α ,β are...

Question

If $α, β$ are roots of $a x^{2} + b x + c = 0$ and $α_{1,} - β$ are roots of $a_{1} x^{2} + b_{1} x + c_{1} = 0$ then find the equation whose roots are $α, α_{1}$

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Solution

$a x^{2} + b x + c = 0$
Let $α$ and $β$ be the roots of $a x^{2} + b x + c = 0$
$α + β = - \frac{b}{a}$ and $α β = \frac{c}{a}$
Let $α_{1}, β$ be the roots of $a_{1} x^{2} + b_{1} x + c_{1} = 0$
$α_{1} + β = - \frac{b_{1}}{a_{1}}$ and $α_{1} β = \frac{c_{1}}{a_{1}}$
Now, $α_{1} + β - α - β = - \frac{b_{1}}{a_{1}} - \frac{b}{a}$
$α_{1} - α = - \frac{b_{1}}{a_{1}} - \frac{b}{a} = \frac{b_{1} a - b a_{1}}{a_{1} a}$
Again $\frac{α_{1} β}{α β} = \frac{\frac{c_{1}}{a_{1}}}{\frac{c}{a}} = \frac{c_{1} a}{a_{1} c}$
$\Rightarrow α_{1} = \frac{c_{1} a}{a_{1} c} α$
Substitute $α_{1} = \frac{c_{1} a}{a_{1} c} α$ in $α_{1} - α = \frac{b_{1} a - b a_{1}}{a_{1} a}$
$\Rightarrow \frac{c_{1} a}{a_{1} c} α - α = \frac{b_{1} a - b a_{1}}{a_{1} a}$
$\Rightarrow (\frac{c_{1} a}{a_{1} c} - 1) α = \frac{b_{1} a - b a_{1}}{a_{1} a}$
$\Rightarrow (\frac{c_{1} a - a_{1} c}{a_{1} c}) α = \frac{b_{1} a - b a_{1}}{a_{1} a}$
$\Rightarrow α = \frac{b_{1} a - b a_{1}}{a_{1} a} \times \frac{a_{1} c}{c_{1} a - a_{1} c}$
$\Rightarrow α = \frac{c (b_{1} a - b a_{1})}{a (c_{1} a - a_{1} c)}$
$∴ α = \frac{c (b_{1} a - b a_{1})}{a (c_{1} a - a_{1} c)}$
We know that $α_{1} = \frac{c_{1} a}{a_{1} c} α$
$\Rightarrow α_{1} = \frac{c_{1} a}{a_{1} c} \times \frac{c (b_{1} a - b a_{1})}{a (c_{1} a - a_{1} c)}$
$\Rightarrow α_{1} = \frac{c_{1} (b_{1} a - b a_{1})}{a_{1} (c_{1} a - a_{1} c)}$
Required equation is $x^{2} - (α + α_{1}) x + α α_{1} = 0$
$x^{2} - (\frac{c (b_{1} a - b a_{1})}{a (c_{1} a - a_{1} c)} + \frac{c_{1} (b_{1} a - b a_{1})}{a_{1} (c_{1} a - a_{1} c)}) x + \frac{c (b_{1} a - b a_{1})}{a (c_{1} a - a_{1} c)} \frac{c_{1} (b_{1} a - b a_{1})}{a_{1} (c_{1} a - a_{1} c)} = 0$
$a a_{1} {(c_{1} a - a_{1} c)}^{2} x^{2} - (a a_{1} (c_{1} a - a_{1} c) (c a_{1} b_{1} a - c {a_{1}}^{2} b + c_{1} b_{1} a^{2} - c_{1} a_{1} a b)) x + c c_{1} {(b_{1} a - b a_{1})}^{2} = 0$
$a a_{1} {(c_{1} a - a_{1} c)}^{2} x^{2} + (a a_{1} (c_{1} a - a_{1} c) (c {a_{1}}^{2} b - c_{1} b_{1} a^{2})) x + c c_{1} {(b_{1} a - b a_{1})}^{2} = 0$ is the required equation.

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