Question

The value of k for which the roots are real and equal of the following equation (k + 1)$x^2$ + 2(k + 3)x + (k + 8) = 0 is $k=\frac{1}{3}$

• A. True
• B. False

Answer 1

The correct option is A True

$\Rightarrow$ The given quadratic equation is $(k+1)x^2+2(k+3)x+(k+8)=0$, comparing it with $a{x}^2+bx+c=0$.

$\Rightarrow$ We get, $a=(k+1),b=2(k+3)$ and $c=k+8$

$\Rightarrow$ The roots are real and equal.

$\therefore$ $b^2-4ac=0$

$\Rightarrow$ $\left(2\right(k+3){)}^2-4(k+1\left)\right(k+8)=0$

$\Rightarrow$ $4(k^2+6k+9)-4(k^2+8k+k+8)=0$

$\Rightarrow$ $4k^2+24k+36-4k^2-36k-32=0$

$\Rightarrow$ $-12k+4=0$

$\Rightarrow$ $12k=4$

$\Rightarrow$ $k=\frac{4}{12}$

$\Rightarrow$ $k=\frac{1}{3}$

$\therefore$ The given statement is true.