Question

If $\alpha ,\beta$ are the roots of the quadratic equation $x^2+3x+6=0$ , then find the equation whose roots are $\frac{1+\alpha }{1-\alpha }$, $\frac{1+\beta }{1-\beta }$

• A. $5y^2+5y+2=0$
• B. $5y^2+2y+5=0$
• C. $2y^2+5y+5=0$
• D. $5y^2-2y+5=0$

Answer 1

The correct option is A

$5y^2+5y+2=0$

Let $x$ is a root of the equation $x^2+3x+6=0$ . We want to find the equation whose roots are $\frac{1+x}{1-x}$
(Because the roots are $\frac{1+\alpha }{1-\alpha }$, $\frac{1+\beta }{1-\beta }$

Let it be $y$.

$\frac{y}{1}=\frac{1+x}{1-x}$

$\frac{y-1}{y+1}=\frac{1+x-(1-x)}{1+x+1-x}=x$

Substitute $\frac{y-1}{y+1}$ in the equation

$x^2+3x+6$ to get the equation whose root is y.

$\Rightarrow {\left(\frac{y-1}{y+1}\right)}^2+3\left(\frac{y-1}{y+1}\right)+6=0\Rightarrow \frac{(y-1{)}^2}{(y+1{)}^2}+\frac{3(y-1)(y+1)}{(y+1{)}^2}+\frac{6(y+1{)}^2}{(y+1{)}^2}=0\Rightarrow y^2-2y+1+3(y^2-1)+6(y^2+2y+1)=010y^2+10y+4=0or5y^2+5y+2=0$

(Replace $y$ by $x$, because change of variable does not affect the equation)

We can solve this by finding the sum of the roots and product of the roots also.