Solve for b b^2-4b+4=0

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

Step 1

Factor using the perfect square rule.

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

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

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

$4 b = 2 \cdot b \cdot 2$

Rewrite the polynomial.

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

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

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

Step 2

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

$b - 2 = 0$

Step 3

Add $2$ to both sides of the equation.

$b = 2$