Algebra Examples

f (x) = 4 x^{2} + 2

$f (x) = 4 x^{2} + 2$ ,

f (x) = 4 x + 1

$f (x) = 4 x + 1$

Step 1

Substitute

4 x + 1

$4 x + 1$ for

f (x)

$f (x)$ .

4 x + 1 = 4 x^{2} + 2

$4 x + 1 = 4 x^{2} + 2$

Step 2

Solve for

x

$x$ .

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Step 2.1

Subtract

4 x^{2}

$4 x^{2}$ from both sides of the equation.

4 x + 1 - 4 x^{2} = 2

$4 x + 1 - 4 x^{2} = 2$

Step 2.2

Subtract

2

$2$ from both sides of the equation.

4 x + 1 - 4 x^{2} - 2 = 0

$4 x + 1 - 4 x^{2} - 2 = 0$

Step 2.3

Subtract

2

$2$ from

1

$1$ .

4 x - 4 x^{2} - 1 = 0

$4 x - 4 x^{2} - 1 = 0$

Step 2.4

Factor the left side of the equation.

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Step 2.4.1

Factor

- 1

$- 1$ out of

4 x - 4 x^{2} - 1

$4 x - 4 x^{2} - 1$ .

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Step 2.4.1.1

Reorder

4 x

$4 x$ and

- 4 x^{2}

$- 4 x^{2}$ .

- 4 x^{2} + 4 x - 1 = 0

$- 4 x^{2} + 4 x - 1 = 0$

Step 2.4.1.2

Factor

- 1

$- 1$ out of

- 4 x^{2}

$- 4 x^{2}$ .

- (4 x^{2}) + 4 x - 1 = 0

$- (4 x^{2}) + 4 x - 1 = 0$

Step 2.4.1.3

Factor

- 1

$- 1$ out of

4 x

$4 x$ .

- (4 x^{2}) - (- 4 x) - 1 = 0

$- (4 x^{2}) - (- 4 x) - 1 = 0$

Step 2.4.1.4

Rewrite

- 1

$- 1$ as

- 1 (1)

$- 1 (1)$ .

- (4 x^{2}) - (- 4 x) - 1 \cdot 1 = 0

$- (4 x^{2}) - (- 4 x) - 1 \cdot 1 = 0$

Step 2.4.1.5

Factor

- 1

$- 1$ out of

- (4 x^{2}) - (- 4 x)

$- (4 x^{2}) - (- 4 x)$ .

- (4 x^{2} - 4 x) - 1 \cdot 1 = 0

$- (4 x^{2} - 4 x) - 1 \cdot 1 = 0$

Step 2.4.1.6

Factor

- 1

$- 1$ out of

- (4 x^{2} - 4 x) - 1 (1)

$- (4 x^{2} - 4 x) - 1 (1)$ .

- (4 x^{2} - 4 x + 1) = 0

$- (4 x^{2} - 4 x + 1) = 0$

- (4 x^{2} - 4 x + 1) = 0

$- (4 x^{2} - 4 x + 1) = 0$

Step 2.4.2

Factor using the perfect square rule.

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Step 2.4.2.1

Rewrite

4 x^{2}

$4 x^{2}$ as

{(2 x)}^{2}

${(2 x)}^{2}$ .

- ({(2 x)}^{2} - 4 x + 1) = 0

$- ({(2 x)}^{2} - 4 x + 1) = 0$

Step 2.4.2.2

Rewrite

1

$1$ as

1^{2}

$1^{2}$ .

- ({(2 x)}^{2} - 4 x + 1^{2}) = 0

$- ({(2 x)}^{2} - 4 x + 1^{2}) = 0$

Step 2.4.2.3

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 (2 x) \cdot 1

$4 x = 2 \cdot (2 x) \cdot 1$

Step 2.4.2.4

Rewrite the polynomial.

- ({(2 x)}^{2} - 2 \cdot (2 x) \cdot 1 + 1^{2}) = 0

$- ({(2 x)}^{2} - 2 \cdot (2 x) \cdot 1 + 1^{2}) = 0$

Step 2.4.2.5

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 = 2 x

$a = 2 x$ and

b = 1

$b = 1$ .

- {(2 x - 1)}^{2} = 0

$- {(2 x - 1)}^{2} = 0$

- {(2 x - 1)}^{2} = 0

$- {(2 x - 1)}^{2} = 0$

- {(2 x - 1)}^{2} = 0

$- {(2 x - 1)}^{2} = 0$

Step 2.5

Divide each term in

- {(2 x - 1)}^{2} = 0

$- {(2 x - 1)}^{2} = 0$ by

- 1

$- 1$ and simplify.

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Step 2.5.1

Divide each term in

- {(2 x - 1)}^{2} = 0

$- {(2 x - 1)}^{2} = 0$ by

- 1

$- 1$ .

\frac{- {(2 x - 1)}^{2}}{- 1} = \frac{0}{- 1}

$\frac{- {(2 x - 1)}^{2}}{- 1} = \frac{0}{- 1}$

Step 2.5.2

Simplify the left side.

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Step 2.5.2.1

Dividing two negative values results in a positive value.

\frac{{(2 x - 1)}^{2}}{1} = \frac{0}{- 1}

$\frac{{(2 x - 1)}^{2}}{1} = \frac{0}{- 1}$

Step 2.5.2.2

Divide

{(2 x - 1)}^{2}

${(2 x - 1)}^{2}$ by

1

$1$ .

{(2 x - 1)}^{2} = \frac{0}{- 1}

${(2 x - 1)}^{2} = \frac{0}{- 1}$

{(2 x - 1)}^{2} = \frac{0}{- 1}

${(2 x - 1)}^{2} = \frac{0}{- 1}$

Step 2.5.3

Simplify the right side.

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Step 2.5.3.1

Divide

0

$0$ by

- 1

$- 1$ .

{(2 x - 1)}^{2} = 0

${(2 x - 1)}^{2} = 0$

{(2 x - 1)}^{2} = 0

${(2 x - 1)}^{2} = 0$

{(2 x - 1)}^{2} = 0

${(2 x - 1)}^{2} = 0$

Step 2.6

Set the

2 x - 1

$2 x - 1$ equal to

0

$0$ .

2 x - 1 = 0

$2 x - 1 = 0$

Step 2.7

Solve for

x

$x$ .

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Step 2.7.1

Add

1

$1$ to both sides of the equation.

2 x = 1

$2 x = 1$

Step 2.7.2

Divide each term in

$2 x = 1$ by

$2$ and simplify.

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Step 2.7.2.1

Divide each term in

$2 x = 1$ by

$2$ .

$\frac{2 x}{2} = \frac{1}{2}$

Step 2.7.2.2

Simplify the left side.

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Step 2.7.2.2.1

Cancel the common factor of

$2$ .

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Step 2.7.2.2.1.1

Cancel the common factor.

$\frac{2 x}{2} = \frac{1}{2}$

Step 2.7.2.2.1.2

Divide

$x$ by

$1$ .

$x = \frac{1}{2}$

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