Question

If $\alpha ,\beta ,\gamma$ be the roots of the equation $a{x}^3+b{x}^2+cx+d=0$. To obtain the equation whose roots are $f(\alpha )$, $f(\beta ),f(\gamma )$, where $f$ is a function, we put $y=f(\alpha )$ and simplify it to obtain $\alpha =g(y)$ (some function of $y)$. Now, $\alpha$ is a root of the equation $a{x}^3+b{x}^2+cx+d=0$, then we obtain the desired equation which is $a\{g(y){\}}^3+b\{g(y){\}}^2+c\{g(y)\}+d=0$
For example, if $\alpha ,\beta ,\gamma$ are the roots of $a{x}^3+b{x}^2+cx+d=0$. To find equation whose roots are $\frac{1}{\alpha },\frac{1}{\beta },\frac{1}{\gamma }$ we put $y=\frac{1}{\alpha }\Rightarrow \alpha =\frac{1}{y}$
As $\alpha$ is a root of $a{x}^3+b{x}^2+cx+d=0$
we get $\frac{a}{y^3}+\frac{b}{y^2}+\frac{c}{y}+d=0\Rightarrow d{y}^3+c{y}^2+by+a=0$
This is desired equation.
If $\alpha ,\beta$ are the roots of the equation $a{x}^2+bx+c=0$, then the roots of the equation $a(2x+1{)}^2+b(2x+1)(x-1)+c(x-1{)}^2=0$ are-

• A. $\frac{2\alpha +1}{\alpha -1},\frac{2\beta +1}{\beta -1}$
• B. $\frac{2\alpha -1}{\alpha +1},\frac{2\beta -1}{\beta +1}$
• C. $\frac{\alpha +1}{\alpha -2},\frac{\beta +1}{\beta -2}$
• D. $\frac{2\alpha +3}{\alpha -1},\frac{2\beta +3}{\beta -1}$

Answer 1

Answer: (C)

$a{\left(\frac{2x+1}{x-1}\right)}^2+b\left(\frac{2x+1}{x-1}\right)+c=0$
$\text{ Let }\frac{2x+1}{x-1}=\alpha$
$\Rightarrow x\alpha -\alpha =2x+1$
$\Rightarrow x(\alpha -2)=1+\alpha \Rightarrow x=\frac{1+\alpha }{\alpha -2}$
Similarly other root can be find out.