Question

Let $\alpha ,\beta$ be the roots of $a{x}^2+bx+c=0$. The roots of $a(x-2{)}^2-b(x-2\left)\right(x-3)+c(x-3{)}^2=0$, where $a\ne 0$ are

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

Answer 1

The correct option is D $\frac{2+3\alpha }{1+\alpha },\frac{2+3\beta }{1+\beta }$

$a{x}^2+bx+c=0\cdots \left(1\right)$
$a(x-2{)}^2-b(x-2\left)\right(x-3)+c(x-3{)}^2=0$
Putting $x=3$
$a\times 1^2=0\Rightarrow a=0$
Which is not possible, so $(x-3)$ is not the root of the equation.
Let the roots of the equation $a(x-2{)}^2-b(x-2\left)\right(x-3)+c(x-3{)}^2=0$ be $x_1,x_2$
Rewritting the equation as,
$(x-3{)}^2[a{\{-\left(\frac{x-2}{x-3}\right)\}}^2+b\{-\left(\frac{x-2}{x-3}\right)\}+c]=0\cdots (2)$
Now, compairing from equation $\left(1\right)$, we can say that,
$-\left(\frac{x_1-2}{x_1-3}\right)=\alpha -\left(\frac{x_2-2}{x_2-3}\right)=\beta$
Solving anyone from them,
$-\left(\frac{x_1-2}{x_1-3}\right)=\alpha \Rightarrow \frac{x_1-3+1}{x_1-3}=-\alpha \Rightarrow 1+\frac{1}{x_1-3}=-\alpha \Rightarrow x_1-3=-\left(\frac{1}{1+\alpha }\right)\Rightarrow x_1=3-\frac{1}{1+\alpha }=\frac{2+3\alpha }{1+\alpha }$
Similarly,
$x_2=\frac{2+3\beta }{1+\beta }$

Hence, the roots of the equation are,
$\frac{2+3\alpha }{1+\alpha },\frac{2+3\beta }{1+\beta }$