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Find the value of $k$ if the equation $\left( {k + 1} \right){x^2} - 2\left( {k - 1} \right)x + 1 = 0$ has real and equal roots.
(This question has multiple correct options)
A) $k = 0$
B) $k = 2$
C) $k = 3$
D) $k = - 2$

seo-qna
Last updated date: 20th Jun 2024
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Answer
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Hint:
In this question we have to find the value of $k$ if the given equation has equal roots.
If in a quadratic equation if roots are real and distinct then \[{b^2} - 4ac > 0\],
If in a quadratic equation if the roots are real and equal then \[{b^2} - 4ac = 0\] and
If in a quadratic equation if the roots are imaginary then \[{b^2} - 4ac < 0\].
Here, we have given the roots are equal and we can find the value of $k$ by applying the above formula in accordance with the given condition.

Complete step by step solution:
We are given a quadratic equation i.e. $\left( {k + 1} \right){x^2} - 2\left( {k - 1} \right)x + 1 = 0$ and the roots of this equation are equal. We have to find the value of $k$.
As we know that the roots of the equation can be obtained for a quadratic equation $a{x^2} + bx + c = 0$ by the formula given below:
$ \Rightarrow x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
Here, by comparing it with given quadratic equation i.e. $\left( {k + 1} \right){x^2} - 2\left( {k - 1} \right)x + 1 = 0$ we have,
$a = k + 1$, $b = - 2\left( {k - 1} \right)$ and $c = 1$ ……..(1)
If the roots are equal in the given quadratic equation then
\[ \Rightarrow {b^2} - 4ac = 0\]
By putting the values of a, b and c from (1), we get,
\[ \Rightarrow {\left[ { - 2\left( {k - 1} \right)} \right]^2} - 4(k + 1)(1) = 0\] ………..(2)
We can find the square of $k - 1$ by applying the formula i.e. ${\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab$. Therefore,
$ \Rightarrow {\left( {k - 1} \right)^2} = {k^2} + 1 - 2k$
$ \Rightarrow {\left[ { - 2\left( {k - 1} \right)} \right]^2} = 4\left( {{k^2} + 1 - 2k} \right)$
Here, when we open the bracket then the square of negative becomes positive and the square of 2 is 4 which is with the square of $k - 1$.
By putting this value in equation (2), we get,
\[ \Rightarrow 4\left( {{k^2} + 1 - 2k} \right) - 4(k + 1)(1) = 0\]
By opening the brackets, we get,
\[ \Rightarrow 4{k^2} + 4 - 8k - 4k - 4 = 0\]
By combining the terms containing $k$ and cancelling 4 with negative 4, we get,
\[ \Rightarrow 4{k^2} - 12k = 0\]
Taking $4k$ common from both terms, we get,
\[ \Rightarrow 4k\left( {k - 3} \right) = 0\]
Therefore, $k = 0$ or $k - 3 = 0$
Taking constant to other side, we get $k = 3$

Hence, $k = 0$ or $k = 3$
Option A and Option C are the required answer.


Note:
We may solve this question by obtaining both the roots and then equate them if students forget the above approach. We know the roots of a quadratic equation are given by:
$
   \Rightarrow \alpha = \dfrac{{ - b + \sqrt {{b^2} - 4ac} }}{{2a}} \\
   \Rightarrow \beta = \dfrac{{ - b - \sqrt {{b^2} - 4ac} }}{{2a}} \\
 $
If both the roots are equal then
$ \Rightarrow \dfrac{{ - b + \sqrt {{b^2} - 4ac} }}{{2a}} = \dfrac{{ - b - \sqrt {{b^2} - 4ac} }}{{2a}}$
Cancelling 2a from both sides, we get,
$ \Rightarrow - b + \sqrt {{b^2} - 4ac} = - b - \sqrt {{b^2} - 4ac} $
Now cancel -b from both sides, we get,
$ \Rightarrow \sqrt {{b^2} - 4ac} = - \sqrt {{b^2} - 4ac} $
$
   \Rightarrow 2\sqrt {{b^2} - 4ac} = 0 \\
   \Rightarrow \sqrt {{b^2} - 4ac} = 0 \\
 $
By squaring both sides, we get,
$ \Rightarrow {b^2} - 4ac = 0$
As you can see that the formula we used in the solution is the same as above. Now, we can find the value of $k$.