Question

Find the non zero value of k for which the quadratic equation $ k{{x}^{2}}+1-2\left( k-1 \right)x+{{x}^{2}}=0$ has equal roots. Hence, find the roots of the equation.

Answer 1

Hint: In order to solve this question, you need to know the formula of the discriminant of the quadratic expression. The formula of the discriminant of the quadratic expression is given by $ {{b}^{2}}-4ac$. If $ {{b}^{2}}-4ac$ is equal to 0, then the quadratic expression has equal roots. So, using this, we can find the value of k and then you can use the quadratic formula to find the roots or you can use factorisation.

Complete step by step solution:
First thing we need to do is to equate the discriminant to 0. The discriminant of the quadratic expression is given by $ {{b}^{2}}-4ac$. If $ {{b}^{2}}-4ac$ is equal to 0, then the quadratic expression has equal roots.
Given a quadratic equation is $ k{{x}^{2}}+1-2\left( k-1 \right)x+{{x}^{2}}=0$. This can be rewritten as:
$ \Rightarrow \left( k+1 \right){{x}^{2}}-2\left( k-1 \right)x+1=0$
From here we can tell that,
$ a=k+1,b=-2\left( k-1 \right),c=1$
Therefore, substituting them in the discriminant formula, we get
$ \Rightarrow {{b}^{2}}-4ac=0$
$ \Rightarrow {{\left( -2\left( k-1 \right) \right)}^{2}}-4\left( k+1 \right)1=0$
$ \Rightarrow 4{{k}^{2}}-8k+4-4k-4=0$
$ \Rightarrow 4{{k}^{2}}-12k=0$
$ \Rightarrow k=0,k=3$
Since they asked for non zero value, we take k as 3.
If k=3, we get.
$ \Rightarrow 4{{x}^{2}}-4x+1=0$
$ \Rightarrow {{\left( 2x-1 \right)}^{2}}=0$
$ \Rightarrow x=\dfrac{1}{2}$

Therefore, we get the final answer of the question as: The value of k is 3 and the value of the two equal roots are $ x=\dfrac{1}{2}$.

Note: To do this question, you need to know the discriminant formula. Without it, you will not be able to solve the question. You should also know the three conditions. If the discriminant is greater than 0, then the polynomial has two real roots. If it is less than zero, then it has complex imaginary roots. If it is equal to 0, then it has two equal roots.