Solve the Inequality for x x^2+4<=4x

$x^{2} + 4 \leq 4 x$

Subtract $4 x$ from both sides of the inequality.

$x^{2} + 4 - 4 x \leq 0$

Convert the inequality to an equation.

$x^{2} + 4 - 4 x = 0$

Factor using the perfect square rule.

Rearrange terms.

$x^{2} - 4 x + 4 = 0$

Rewrite $4$ as $2^{2}$ .

$x^{2} - 4 x + 2^{2} = 0$

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$4 x = 2 \cdot x \cdot 2$

Rewrite the polynomial.

$x^{2} - 2 \cdot x \cdot 2 + 2^{2} = 0$

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = x$ and $b = 2$ .

${(x - 2)}^{2} = 0$

Set the $x - 2$ equal to $0$ .

$x - 2 = 0$

Add $2$ to both sides of the equation.

$x = 2$