Question

If $\alpha$ and $\beta$ are the roots of the equation $x^2-2x+3=0$. Find the equation whose roots are ${\displaystyle \frac{\alpha -1}{\alpha +1}},{\displaystyle \frac{\beta -1}{\beta +1}}$

a) $3x^2-2x+1=0$

b) $x^2-x+3=0$

c) $5x^2-2x+3=0$ 

d) $x^2-2x+7=0$

Answer 1

Hint: For the given quadratic equation first find the sum of roots ($\alpha +\beta$) and product of roots ($\alpha \beta$). Now using this find the sum of roots (${\displaystyle \frac{\alpha -1}{\alpha +1}}+{\displaystyle \frac{\beta -1}{\beta +1}}$) and product of roots (${\displaystyle \frac{\alpha -1}{\alpha +1}}{\displaystyle \frac{\beta -1}{\beta +1}}$) of the second quadratic equation and substitute these values in standard form of quadratic equation to find the required solution.

Complete step by step answer:

It is given that $\alpha$ and $\beta$ are the roots of the quadratic equation $x^2-2x+3=0$...............(1)

So using the concept of sum of roots $\alpha +\beta ={\displaystyle \frac{-b}{a}}$ and product of roots $\alpha \beta ={\displaystyle \frac{c}{a}}$.

Comparing equation (1) with standard form of quadratic equation i.e. $a{x}^2+bx+c=0$ we get a=1, b=-2 and c=3.

$\therefore$ We get $\alpha +\beta =2$ and $\alpha \beta =3$

Now we have to find equation with roots ${\displaystyle \frac{\alpha -1}{\alpha +1}},{\displaystyle \frac{\beta -1}{\beta +1}}$

So firstly lets calculate ${\displaystyle \frac{\alpha -1}{\alpha +1}}+{\displaystyle \frac{\beta -1}{\beta +1}}={\displaystyle \frac{\left(\alpha -1\right)\left(\beta +1\right)+\left(\alpha +1\right)\left(\beta -1\right)}{\left(\alpha +1\right)\left(\beta +1\right)}}$

Lets simplify the numerator part,

$\Rightarrow {\displaystyle \frac{\alpha \beta +\alpha -\beta -1+\alpha \beta -\alpha +\beta -1}{\alpha \beta +\alpha +\beta +1}}={\displaystyle \frac{2\alpha \beta -2}{\alpha \beta +\alpha +\beta +1}}$

Now substituting the values of $\alpha \beta$ and $\alpha +\beta$

We have ${\displaystyle \frac{6-2}{3+2+1}}={\displaystyle \frac{4}{6}}={\displaystyle \frac{2}{3}}$.............(2)

Now in the similar way we will be computing the value of 

$\Rightarrow \left({\displaystyle \frac{\alpha -1}{\alpha +1}}\right)\left({\displaystyle \frac{\beta -1}{\beta +1}}\right)={\displaystyle \frac{\alpha \beta -\alpha -\beta +1}{\alpha \beta +\alpha +\beta +1}}$

Again substituting the values of $\alpha$ and $\beta$, we have 

$\Rightarrow {\displaystyle \frac{3-2+1}{3+2+1}}={\displaystyle \frac{2}{6}}={\displaystyle \frac{1}{3}}$................(3)

Now using the concept that we can write a quadratic equation using some of roots and products of roots which is $\Rightarrow x^2$ - (sum of roots)x +(products of roots) = 0 

Hence the equation having roots ${\displaystyle \frac{\alpha -1}{\alpha +1}}$, ${\displaystyle \frac{\beta -1}{\beta +1}}$ is

$\Rightarrow x^2-{\displaystyle \frac{2}{3}}x+{\displaystyle \frac{1}{3}}=0$

On simplification we get $3x^2-2x+1=0$

Hence option (a) is the right answer.

Note - In such types of problems always use the concept of sum of roots and product of roots and use this to obtain the quadratic equation.