Question

The smallest integer value of $k$ for which roots of the equation $x^2-8kx+16(k^2-k+1)=0$ are real is

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is A $1$

Given that,

$x^2-8kx+16(k^2-k+1)=0$ has real roots.

To find out,

The smallest integer value of $k$

We know that, the discriminant of a quadratic equation of the form $a{x}^2+bx+c=0$ is given by,

$D=b^2-4ac$

Also, for an equation to have real roots, $D=0$

Hence, $b^2-4ac=0$

$\Rightarrow b^2=4ac$

Here, $a=1,b=8k$ and $c=16(k^2-k+1)$

Hence, $(8k{)}^2=4\times 1\times 16(k^2-k+1)$

$\Rightarrow 64k^2=64(k^2-k+1)$

$\Rightarrow k^2=k^2-k+1$

$\Rightarrow k=1$

Hence, the smallest integer value of $k$ is $1$.