Question

Find the values of k for which the quadratic equation has equal roots. Also, find the roots.

Answer 1

The given quadric equation is $\left(3k+1\right)x^2+2\left(k+1\right)x+1=0$, and roots are real and equal.

Then, find the value of k.

Here, $a=3k+1,b=2(k+1)\mathrm{and}c=1$.

As we know that $D=b^2-4ac$

Putting the values of $a=3k+1,b=2(k+1)\mathrm{and}c=1$.

$$\begin{gathered} D={\left[2\left(k+1\right)\right]}^2-4\left(3k+1\right)\left(1\right) \\ =4(k^2+2k+1)-12k-4 \\ =4k^2+8k+4-12k-4 \\ =4k^2-4k \end{gathered}$$

The given equation will have real and equal roots, if D = 0

Thus, $4k^2-4k=0$
$$\begin{gathered} \Rightarrow 4k(k-1)=0 \\ \Rightarrow k=0\mathrm{or}k-1=0 \\ \Rightarrow k=0\mathrm{or}k=1 \end{gathered}$$

Therefore, the value of k is 0 or 1.

Now, for k = 0, the equation becomes

$$\begin{gathered} x^2+2x+1=0 \\ \Rightarrow x^2+x+x+1=0 \\ \Rightarrow x(x+1)+1(x+1)=0 \\ \Rightarrow (x+1{)}^2=0 \\ \Rightarrow x=-1,-1 \end{gathered}$$

for k = 1, the equation becomes

$$\begin{gathered} 4x^2+4x+1=0 \\ \Rightarrow 4x^2+2x+2x+1=0 \\ \Rightarrow 2x(2x+1)+1(2x+1)=0 \\ \Rightarrow (2x+1{)}^2=0 \\ \Rightarrow x=-\frac{1}{2},-\frac{1}{2} \end{gathered}$$

Hence, the roots of the equation are $-1\mathrm{and}-\frac{1}{2}$.