Question

For what value of k does the quadratic equation ${{x}^{2}}-2(k+1)x+{{k}^{2}}=0$ have equal and real roots?

Answer 1

As we know that a quadratic equation has equal and real roots when the value of discriminant is equal to zero. So by using this concept we will find the value of k. The discriminant of a quadratic equation of the form $a{{x}^{2}}+bx+c=0$ is given as $D=\sqrt{{{b}^{2}}-4ac}$.

Complete step by step solution:
We have been given an equation ${{x}^{2}}-2(k+1)x+{{k}^{2}}=0$.
We have to find the value of k for which the given equation has equal and real roots.
Now, we know that the general form of a quadratic equation is $a{{x}^{2}}+bx+c=0$. Now, by comparing the given equation with the general equation we get the values as
$a=1,b=2(k+1),c={{k}^{2}}$
The equation has equal and real roots when the discriminant is zero.
The discriminant of a quadratic equation is given as $D=\sqrt{{{b}^{2}}-4ac}$.
Now, substituting the values in the above formula we will get
$\begin{aligned} & \Rightarrow D=0 \\\ & \Rightarrow \sqrt{{{b}^{2}}-4ac}=0 \\\ & \Rightarrow {{b}^{2}}-4ac=0 \\\ & \Rightarrow {{2}^{2}}{{\left( k+1 \right)}^{2}}-4\times 1\times {{k}^{2}}=0 \\\ \end{aligned}$
Now, simplifying the above obtained equation we will get
$\begin{aligned} & \Rightarrow 4\left( {{k}^{2}}+2k+1 \right)-4{{k}^{2}}=0 \\\ & \Rightarrow 4{{k}^{2}}+8k+4-4{{k}^{2}}=0 \\\ \end{aligned}$
Now, simplifying further we will get
$\begin{aligned} & \Rightarrow 8k+4=0 \\\ & \Rightarrow 8k=-4 \\\ & \Rightarrow k=\dfrac{-4}{8} \\\ & \Rightarrow k=\dfrac{-1}{2} \\\ \end{aligned}$
Hence the given equation has equal and real roots when the value of k is $\dfrac{-1}{2}$.

Note: The point to be remembered is that the number of roots of a quadratic equation depends on the degree of the equation. In this particular question the degree of the equation is two. We can also find the roots of the equation by using the formula $x=\dfrac{-b}{2a}$ when the value of discriminant is zero.