Algebra Examples

$16 x^{2} - 9 = 0$

Step 1

Rewrite $16 x^{2}$ as ${(4 x)}^{2}$ .

${(4 x)}^{2} - 9 = 0$

Step 2

Rewrite $9$ as $3^{2}$ .

${(4 x)}^{2} - 3^{2} = 0$

Step 3

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 4 x$ and $b = 3$ .

$(4 x + 3) (4 x - 3) = 0$

Step 4

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$4 x + 3 = 0$

$4 x - 3 = 0$

Step 5

Set $4 x + 3$ equal to $0$ and solve for $x$ .

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Step 5.1

Set $4 x + 3$ equal to $0$ .

$4 x + 3 = 0$

Step 5.2

Solve $4 x + 3 = 0$ for $x$ .

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Step 5.2.1

Subtract $3$ from both sides of the equation.

$4 x = - 3$

Step 5.2.2

Divide each term in $4 x = - 3$ by $4$ and simplify.

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Step 5.2.2.1

Divide each term in $4 x = - 3$ by $4$ .

$\frac{4 x}{4} = \frac{- 3}{4}$

Step 5.2.2.2

Simplify the left side.

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Step 5.2.2.2.1

Cancel the common factor of $4$ .

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Step 5.2.2.2.1.1

Cancel the common factor.

$\frac{4 x}{4} = \frac{- 3}{4}$

Step 5.2.2.2.1.2

Divide $x$ by $1$ .

$x = \frac{- 3}{4}$

Step 5.2.2.3

Simplify the right side.

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Step 5.2.2.3.1

Move the negative in front of the fraction.

$x = - \frac{3}{4}$

Step 6

Set $4 x - 3$ equal to $0$ and solve for $x$ .

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Step 6.1

Set $4 x - 3$ equal to $0$ .

$4 x - 3 = 0$

Step 6.2

Solve $4 x - 3 = 0$ for $x$ .

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Step 6.2.1

Add $3$ to both sides of the equation.

$4 x = 3$

Step 6.2.2

Divide each term in $4 x = 3$ by $4$ and simplify.

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Step 6.2.2.1

Divide each term in $4 x = 3$ by $4$ .

$\frac{4 x}{4} = \frac{3}{4}$

Step 6.2.2.2

Simplify the left side.

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Step 6.2.2.2.1

Cancel the common factor of $4$ .

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Step 6.2.2.2.1.1

Cancel the common factor.

$\frac{4 x}{4} = \frac{3}{4}$

Step 6.2.2.2.1.2

Divide $x$ by $1$ .

$x = \frac{3}{4}$

Step 7

The final solution is all the values that make $(4 x + 3) (4 x - 3) = 0$ true.

$x = - \frac{3}{4}, \frac{3}{4}$