Convert to Interval Notation x^2-8x+16

$x^{2} - 8 x + 16 < 0$

Convert the inequality to an equation.

$x^{2} - 8 x + 16 = 0$

Factor using the perfect square rule.

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Rewrite $16$ as $4^{2}$ .

$x^{2} - 8 x + 4^{2} = 0$

Check the middle term by multiplying $2 a b$ and compare this result with the middle term in the original expression.

$2 a b = 2 \cdot x \cdot - 4$

Simplify.

$2 a b = - 8 x$

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = x$ and $b = - 4$ .

${(x - 4)}^{2} = 0$

Set the $x - 4$ equal to $0$ .

$x - 4 = 0$

Add $4$ to both sides of the equation.

$x = 4$

Convert the inequality to interval notation.

$[4, 4]$

Convert to Interval Notation x^2-8x+16<0

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