Question

If the roots of the equation $a{x}^2-bx+c=0$ are $\alpha ,\beta$, then the roots of the equation $b^{2}c{x}^2-a{b}^{2}x+a^3=0$ are

• A. $\frac{1}{{\alpha }^3+\alpha \beta },\frac{1}{{\beta }^3+\alpha \beta }$
• B. $\frac{1}{{\alpha }^2+\alpha \beta },\frac{1}{{\beta }^2+\alpha \beta }$
• C. $\frac{1}{{\alpha }^4+\alpha \beta },\frac{1}{{\beta }^4+\alpha \beta }$
• D. none of these

Answer 1

Answer: (B)

$\frac{1}{{\alpha }^2+\alpha \beta },\frac{1}{{\beta }^2+\alpha \beta }$

Given: $\alpha ,\beta$ are roots of the equation $a{x}^2-bx+c=0$

Solution:

From the equation $a{x}^2-bx+c=0$

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

Now the second equation $b^{2}c{x}^2-a{b}^{2}x+a^3=0$ is of the form $a{x}^2+bx+c=0$ where $a=b^{2}c,b=-a{b}^2,c=a^3$,

and its roots can be written as

$\Rightarrow x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$\Rightarrow x=\frac{a{b}^2\pm \sqrt{(-a{b}^2{)}^2-4(b^{2}c\left)\right(a^3)}}{2\left(b^{2}c\right)}$

$\Rightarrow x=\frac{a{b}^2\pm \sqrt{a^2{b}^4-4a^3{b}^{2}c}}{2\left(b^{2}c\right)}$

Divide the above equation throughout with $a^{2}b$, we get

$x=\frac{\frac{a{b}^2}{a^{2}b}\pm \frac{1}{a^{2}b}\sqrt{a^2{b}^4-4a^3{b}^{2}c}}{\frac{2b^{2}c}{a^{2}b}}$

$\Rightarrow x=\frac{\frac{b}{a}\pm \sqrt{\frac{a^2{b}^4-4a^3{b}^{2}c}{a^4{b}^2}}}{2\frac{b}{a}\frac{c}{a}}$

$\Rightarrow x=\frac{\frac{b}{a}\pm \sqrt{{\left(\frac{b}{a}\right)}^2-4\frac{c}{a}}}{2\frac{b}{a}\frac{c}{a}}$

Substitute the values of $\frac{b}{a},\frac{c}{a}$, we get

$x=\frac{\alpha +\beta \pm \sqrt{(\alpha +\beta {)}^2-4\alpha \beta }}{2(\alpha +\beta )\left(\alpha \beta \right)}$

$\Rightarrow x=\frac{\alpha +\beta \pm \sqrt{{\alpha }^2+{\beta }^2+2\alpha \beta -4\alpha \beta }}{2(\alpha +\beta )\left(\alpha \beta \right)}$

$\Rightarrow x=\frac{\alpha +\beta \pm \sqrt{{\alpha }^2+{\beta }^2-2\alpha \beta }}{2(\alpha +\beta )\left(\alpha \beta \right)}$

$\Rightarrow x=\frac{\alpha +\beta \pm \sqrt{(\alpha -\beta {)}^2}}{2(\alpha +\beta )\left(\alpha \beta \right)}$

So the roots are

$x=\frac{\alpha +\beta +(\alpha -\beta )}{2(\alpha +\beta )\left(\alpha \beta \right)},x=\frac{\alpha +\beta -(\alpha -\beta )}{2(\alpha +\beta )\left(\alpha \beta \right)}$

$\Rightarrow x=\frac{1}{{\beta }^2+\alpha \beta },\frac{1}{{\alpha }^2+\alpha \beta }$