Find the Roots (Zeros) f(x)=(x^2-4x+4)/(x^2-x-2)

$f (x) = \frac{x^{2} - 4 x + 4}{x^{2} - x - 2}$

Step 1

Set $\frac{x^{2} - 4 x + 4}{x^{2} - x - 2}$ equal to $0$ .

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

Step 2

Solve for $x$ .

Set the numerator equal to zero.

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

Solve the equation for $x$ .

Factor using the perfect square rule.

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.

$x = 2$

Exclude the solutions that do not make $\frac{x^{2} - 4 x + 4}{x^{2} - x - 2} = 0$ true.

$No solution$