Question

If $\alpha ,\beta$ are roots of $a{x}^2+bx+c=0$ and ${\alpha }_{1,}-\beta$ are roots of $a_1{x}^2+b_{1}x+c_1=0$ then find the equation whose roots are $\alpha ,{\alpha }_1$

Answer 1

$a{x}^2+bx+c=0$

Let $\alpha$ and $\beta$ be the roots of $a{x}^2+bx+c=0$

$\alpha +\beta =-\frac{b}{a}$ and $\alpha \beta =\frac{c}{a}$

Let ${\alpha }_1,\beta$ be the roots of $a_1{x}^2+b_{1}x+c_1=0$

${\alpha }_1+\beta =-\frac{b_1}{a_1}$ and ${\alpha }_1\beta =\frac{c_1}{a_1}$

Now, ${\alpha }_1+\beta -\alpha -\beta =-\frac{b_1}{a_1}-\frac{b}{a}$

${\alpha }_1-\alpha =-\frac{b_1}{a_1}-\frac{b}{a}=\frac{b_{1}a-b{a}_1}{a_{1}a}$

Again $\frac{{\alpha }_1\beta }{\alpha \beta }=\frac{\frac{c_1}{a_1}}{\frac{c}{a}}=\frac{c_{1}a}{a_{1}c}$

$\Rightarrow {\alpha }_1=\frac{c_{1}a}{a_{1}c}\alpha$

Substitute ${\alpha }_1=\frac{c_{1}a}{a_{1}c}\alpha$ in ${\alpha }_1-\alpha =\frac{b_{1}a-b{a}_1}{a_{1}a}$

$\Rightarrow \frac{c_{1}a}{a_{1}c}\alpha -\alpha =\frac{b_{1}a-b{a}_1}{a_{1}a}$

$\Rightarrow (\frac{c_{1}a}{a_{1}c}-1)\alpha =\frac{b_{1}a-b{a}_1}{a_{1}a}$

$\Rightarrow \left(\frac{c_{1}a-a_{1}c}{a_{1}c}\right)\alpha =\frac{b_{1}a-b{a}_1}{a_{1}a}$

$\Rightarrow \alpha =\frac{b_{1}a-b{a}_1}{a_{1}a}\times \frac{a_{1}c}{c_{1}a-a_{1}c}$

$\Rightarrow \alpha =\frac{c(b_{1}a-b{a}_1)}{a(c_{1}a-a_{1}c)}$

$\therefore \alpha =\frac{c(b_{1}a-b{a}_1)}{a(c_{1}a-a_{1}c)}$

We know that ${\alpha }_1=\frac{c_{1}a}{a_{1}c}\alpha$

$\Rightarrow {\alpha }_1=\frac{c_{1}a}{a_{1}c}\times \frac{c(b_{1}a-b{a}_1)}{a(c_{1}a-a_{1}c)}$

$\Rightarrow {\alpha }_1=\frac{c_1(b_{1}a-b{a}_1)}{a_1(c_{1}a-a_{1}c)}$

Required equation is $x^2-(\alpha +{\alpha }_1)x+\alpha {\alpha }_1=0$

$x^2-(\frac{c(b_{1}a-b{a}_1)}{a(c_{1}a-a_{1}c)}+\frac{c_1(b_{1}a-b{a}_1)}{a_1(c_{1}a-a_{1}c)})x+\frac{c(b_{1}a-b{a}_1)}{a(c_{1}a-a_{1}c)}\frac{c_1(b_{1}a-b{a}_1)}{a_1(c_{1}a-a_{1}c)}=0$

$a{a}_1{(c_{1}a-a_{1}c)}^2{x}^2-\left(a{a}_1(c_{1}a-a_{1}c)(c{a}_1{b}_{1}a-c{a_1}^{2}b+c_1{b}_1{a}^2-c_1{a}_{1}ab)\right)x+c{c}_1{(b_{1}a-b{a}_1)}^2=0$

$a{a}_1{(c_{1}a-a_{1}c)}^2{x}^2+\left(a{a}_1(c_{1}a-a_{1}c)(c{a_1}^{2}b-c_1{b}_1{a}^2)\right)x+c{c}_1{(b_{1}a-b{a}_1)}^2=0$ is the required equation.