Question

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

• A. True
• B. False

Answer 1

The correct option is A True

$x^2-2(k+1)x+k^2=0-$

$\Rightarrow$ Here, $a=1,b=-2(k+1),c=k^2$

$\Rightarrow$ It is given that roots area real and equal.

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

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

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

$\Rightarrow$ $4(k^2+2k+1)-4k^2=0$

$\Rightarrow$ $4k^2+8k+4-4k^2=0$

$\Rightarrow$ $8k+4=0$

$\Rightarrow$ $8k=-4$

$\therefore$ $k=\frac{-1}{2}$

$\therefore$ We can see, value of $k$ given in question is correct.