Solve for b 6^2+b^2=(3 square root of 5)^2

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Algebra

$6^{2} + b^{2} = {(3 \sqrt{5})}^{2}$

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

$36 + b^{2} = {(3 \sqrt{5})}^{2}$

Simplify ${(3 \sqrt{5})}^{2}$ .

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Simplify the expression.

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Apply the product rule to $3 \sqrt{5}$ .

$36 + b^{2} = 3^{2} {\sqrt{5}}^{2}$

Raise $3$ to the power of $2$ .

$36 + b^{2} = 9 {\sqrt{5}}^{2}$

Rewrite ${\sqrt{5}}^{2}$ as $5$ .

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Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{5}$ as $5^{\frac{1}{2}}$ .

$36 + b^{2} = 9 {(5^{\frac{1}{2}})}^{2}$

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

$36 + b^{2} = 9 \cdot 5^{\frac{1}{2} \cdot 2}$

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

$36 + b^{2} = 9 \cdot 5^{\frac{2}{2}}$

Cancel the common factor of $2$ .

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Cancel the common factor.

$36 + b^{2} = 9 \cdot 5^{\frac{2}{2}}$

Rewrite the expression.

$36 + b^{2} = 9 \cdot 5^{1}$

Evaluate the exponent.

$36 + b^{2} = 9 \cdot 5$

Multiply $9$ by $5$ .

$36 + b^{2} = 45$

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

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Subtract $36$ from both sides of the equation.

$b^{2} = 45 - 36$

Subtract $36$ from $45$ .

$b^{2} = 9$

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

$b = \pm \sqrt{9}$

Simplify $\pm \sqrt{9}$ .

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

$b = \pm \sqrt{3^{2}}$

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

$b = \pm 3$

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

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

$b = 3$

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

$b = - 3$

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

$b = 3, - 3$

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