Question

If the roots of the quadratic equation $\left( {k + 1} \right){x^2} - 2\left( {k - 1} \right)x + 1 = 0$ are real and equal, then find the value of k.

Answer 1

Hint – In this question consider the roots as $\alpha $. Then use the direct formula for the sum of the roots and the product of the roots in terms of coefficients of a quadratic equation of the form $a{x^2} + bx + c = 0$ this will help to get the right answer.
Complete step-by-step answer:
Given quadratic equation is
$\left( {k + 1} \right){x^2} - 2\left( {k - 1} \right)x + 1 = 0$
Now it is given that the roots of this quadratic equation are real and equal (i.e. both the roots are the same).
So let us consider both the roots are $\alpha $.
Now as we know that in a quadratic equation the sum of roots is the ratio of negative time’s coefficient of x to the coefficient of x².
$ \Rightarrow \alpha + \alpha = \dfrac{{ - \left( { - 2\left( {k - 1} \right)} \right)}}{{k + 1}} = \dfrac{{2k - 2}}{{k + 1}}$
$ \Rightarrow 2\alpha = \dfrac{{2\left( {k - 1} \right)}}{{k + 1}}$
$ \Rightarrow \alpha = \dfrac{{k - 1}}{{k + 1}}$……………………… (1)
Now we also know that in a quadratic equation the product of roots is the ratio of constant term to the coefficient of x².
$ \Rightarrow {\alpha ^2} = \dfrac{1}{{k + 1}}$…………………. (2)
Now from equation (1) put the value of $\alpha $ in equation (2) we have,
$ \Rightarrow {\left( {\dfrac{{k - 1}}{{k + 1}}} \right)^2} = \dfrac{1}{{k + 1}}$
Now simplify the above equation we have,
$ \Rightarrow {\left( {k - 1} \right)^2} = \dfrac{{{{\left( {k + 1} \right)}^2}}}{{k + 1}} = k + 1$
Now expand the square according to property ${\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab$ we have,
$ \Rightarrow {k^2} + 1 - 2k = k + 1$
Now simplify the above equation we have,
$ \Rightarrow {k^2} - 3k = 0$
$ \Rightarrow k\left( {k - 3} \right) = 0$
$ \Rightarrow k = 0,3$
So this is the required value of k such that the given quadratic equation has real and equal roots.
So this is the required answer.

Note – There can be another method to solve this problem. If the roots of a quadratic equation are real and equal then the value discriminant that is ${b^2} - 4ac = 0$ for any quadratic equation of the form $a{x^2} + bx + c = 0{\text{ where a}} \ne {\text{0}}$.