Question

Find the least integral value of $k$ for which the equation $x^2-2(k+2)x+12+k^2=0$ has two different real roots.

Answer 1

Given quadratic equation is $x^2-2(k+2)x+12+k^2=0$

For two different real roots, $D>0$

$\Rightarrow b^2-4ac>0$

$\Rightarrow \left(2\right(k+2){)}^2-4\times 1\times (12+k^2)>0$

$\Rightarrow 4(k^2+4+4k)-48-4k^2>0$

$\Rightarrow 4k^2+16+16k-48-4k^2>0$

$\Rightarrow 16k-32>0$

$\Rightarrow k>2$

Hence, the least integral value of $k$ is $3$.