Question

(i) Find the values of k for which the quadratic equation $\left(3k+1\right)x^2+2\left(k+1\right)x+1=0$ has real and equal roots. [CBSE 2014]
(ii) Find the value of k for which the equation $x^2+k\left(2x+k-1\right)+2=0$ has real and equal roots. [CBSE 2017]

Answer 1

(i)
The given equation is $\left(3k+1\right)x^2+2\left(k+1\right)x+1=0$.

This is of the form $a{x}^2+bx+c=0$, where a = 3k +1, b = 2(k + 1) and c = 1.

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

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

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

Hence, 0 and 1 are the required values of k.

(ii)
$$\begin{gathered} x^2+k\left(2x+k-1\right)+2=0 \\ {x}^2+2kx+k^2-k+2=0 \\ \mathrm{For}\mathrm{real}\mathrm{and}\mathrm{equal}\mathrm{roots}D=0 \\ D=b^2-4ac=0 \\ D={\left(2k\right)}^2-4\left(1\right)\left(k^2-k+2\right) \\ D=4k^2-4k^2+4k-8=0 \\ \Rightarrow 4k-8=0 \\ \Rightarrow k=2 \end{gathered}$$