Pre-Algebra Examples

${(6 \sqrt{3})}^{2} + b^{2} = 12^{2}$

Step 1

Simplify each term.

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Step 1.1

Apply the product rule to $6 \sqrt{3}$ .

$6^{2} {\sqrt{3}}^{2} + b^{2} = 12^{2}$

Step 1.2

Raise $6$ to the power of $2$ .

$36 {\sqrt{3}}^{2} + b^{2} = 12^{2}$

Step 1.3

Rewrite ${\sqrt{3}}^{2}$ as $3$ .

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Step 1.3.1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{3}$ as $3^{\frac{1}{2}}$ .

$36 {(3^{\frac{1}{2}})}^{2} + b^{2} = 12^{2}$

Step 1.3.2

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$36 \cdot 3^{\frac{1}{2} \cdot 2} + b^{2} = 12^{2}$

Step 1.3.3

Combine $\frac{1}{2}$ and $2$ .

$36 \cdot 3^{\frac{2}{2}} + b^{2} = 12^{2}$

Step 1.3.4

Cancel the common factor of $2$ .

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Step 1.3.4.1

Cancel the common factor.

$36 \cdot 3^{\frac{2}{2}} + b^{2} = 12^{2}$

Step 1.3.4.2

Rewrite the expression.

$36 \cdot 3^{1} + b^{2} = 12^{2}$

Step 1.3.5

Evaluate the exponent.

$36 \cdot 3 + b^{2} = 12^{2}$

Step 1.4

Multiply $36$ by $3$ .

$108 + b^{2} = 12^{2}$

Step 2

Raise $12$ to the power of $2$ .

$108 + b^{2} = 144$

Step 3

Move all terms not containing $b$ to the right side of the equation.

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Step 3.1

Subtract $108$ from both sides of the equation.

$b^{2} = 144 - 108$

Step 3.2

Subtract $108$ from $144$ .

$b^{2} = 36$

Step 4

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

$b = \pm \sqrt{36}$

Step 5

Simplify $\pm \sqrt{36}$ .

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Step 5.1

Rewrite $36$ as $6^{2}$ .

$b = \pm \sqrt{6^{2}}$

Step 5.2

Pull terms out from under the radical, assuming positive real numbers.

$b = \pm 6$

Step 6

The complete solution is the result of both the positive and negative portions of the solution.

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Step 6.1

First, use the positive value of the $\pm$ to find the first solution.

$b = 6$

Step 6.2

Next, use the negative value of the $\pm$ to find the second solution.

$b = - 6$

Step 6.3

The complete solution is the result of both the positive and negative portions of the solution.

$b = 6, - 6$