Question

Find the values of $k$ for which the equation $(3k+1)x^2+2(k+1)x+1=0$ has equal roots. Also, find the roots.

Answer 1

Step 1: Evaluate the discriminant $D$ for the given equation $(3k+1)x^2+2(k+1)x+1=0$

Compare with $a{x}^2+bx+c=0$

Here, $a=(3k+1),b=2(k+1),c=1$

If an equation has equal roots, then discriminant $D=0$.

$$\begin{gathered} \Rightarrow b^2-4ac=D \\ \Rightarrow 4{(k+1)}^2-4(3k+1)\left(1\right)=0 \\ \Rightarrow 4(k^2+2k+1)-12k-4=0 \\ \Rightarrow 4k^2+8k+4-12k-4=0 \\ \Rightarrow 4k^2-4k=0 \\ \Rightarrow 4k(k-1)=0 \end{gathered}$$

Eithe $4k=0$ or $k-1=0$

So, $k=0$ or $k=1$

Hence, the possible values of $k$ are $0$ or $1$.

Step 2: Find the required equations

Put value of $k$ in the equation

$\Rightarrow$ If $k=0$, then $x^2+2x+1=0$

$\Rightarrow$ If $k=1$, then $4x^2+4x+1=0$

Step 3: Factorise the equations to get the roots

$x^2+2x+1=0$

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

Thus, the roots of the equation when $k=0$ are $-1,-1$.

$4x^2+4x+1=0$

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

Thus, the roots of the equation when $k=1$ are $-\frac{1}{2},-\frac{1}{2}$.

Hence,the values of $k$ are $0$ ,$1$and corresponding roots are $-1,-1$ and $\frac{-1}{2},\frac{-1}{2}$ respectively.