Question

Let $x^2+x+k>0,$ then prove that $k>\frac{1}{4}$

Answer 1

$x^2+x+k>0$

To prove $k>\frac{1}{4}$

$x^2+x+\frac{1}{4}-\frac{1}{4}+k>0$

${(x+\frac{1}{2})}^2-\frac{1}{4}+k>0$

${(x+\frac{1}{2})}^2+(k-\frac{1}{4})>0$

we know that ${(x+\frac{1}{2})}^2$ is always greater then or equal to zero.

so, $(k-\frac{1}{4})$ should also be greater than zero to satisfy the equation

$\Rightarrow k-\frac{1}{4}>0$

$k>\frac{1}{4}$

Hence proved