Algebra Examples

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

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

Step 1

Set

\frac{x^{2} - 4 x + 4}{x^{2} - 4}

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

0

$0$ .

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

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

Step 2

Solve for

x

$x$ .

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Step 2.1

Set the numerator equal to zero.

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

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

Step 2.2

Solve the equation for

x

$x$ .

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Step 2.2.1

Factor using the perfect square rule.

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Step 2.2.1.1

Rewrite

4

$4$ as

2^{2}

$2^{2}$ .

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

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

Step 2.2.1.2

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

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

Step 2.2.1.3

Rewrite the polynomial.

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

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

Step 2.2.1.4

Factor using the perfect square trinomial rule

a^{2} - 2 a b + b^{2} = {(a - b)}^{2}

$a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where

a = x

$a = x$ and

b = 2

$b = 2$ .

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

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

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

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

Step 2.2.2

Set the

x - 2

$x - 2$ equal to

0

$0$ .

x - 2 = 0

$x - 2 = 0$

Step 2.2.3

Add

2

$2$ to both sides of the equation.

x = 2

$x = 2$

x = 2

$x = 2$

Step 2.3

Exclude the solutions that do not make

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

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

No solution

$No solution$

No solution

$No solution$