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# If $\alpha$ and $\beta$ are the roots of the equation $x^2 - 2x + 3 = 0$. Find the equation whose roots are $\dfrac{{\alpha - 1}}{{\alpha + 1}},\dfrac{{\beta - 1}}{{\beta + 1}}$a) $3x^2 - 2x + 1 = 0$b) $x^2 - x + 3 = 0$c) $5x^2 - 2x + 3 = 0$ d) $x^2 - 2x + 7 = 0$ Verified
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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 ($\dfrac{{\alpha - 1}}{{\alpha + 1}} + \dfrac{{\beta - 1}}{{\beta + 1}}$) and product of roots ($\dfrac{{\alpha - 1}}{{\alpha + 1}} \dfrac{{\beta - 1}}{{\beta + 1}}$) of the second quadratic equation and substitute these values in standard form of quadratic equation to find the required solution.

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 $\alpha + \beta = \dfrac{{ - b}}{a}$ and product of roots $\alpha\beta=\dfrac{c}{a}$.

Comparing equation (1) with standard form of quadratic equation i.e. $ax^2+bx+c=0$ 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 $\dfrac{{\alpha - 1}}{{\alpha + 1}},\dfrac{{\beta - 1}}{{\beta + 1}}$

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

Lets simplify the numerator part,

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

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

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

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

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

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

$\Rightarrow \dfrac{{3 - 2 + 1}}{{3 + 2 + 1}} = \dfrac{2}{6} = \dfrac{1}{3}$................(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 $\dfrac{{\alpha - 1}}{{\alpha + 1}}$, $\dfrac{{\beta - 1}}{{\beta + 1}}$ is

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

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.

Last updated date: 17th Sep 2023
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