Algebra Examples

$c = \sqrt{a^{2} + b^{2}}$

Step 1

Rewrite the equation as $\sqrt{a^{2} + b^{2}} = c$ .

$\sqrt{a^{2} + b^{2}} = c$

Step 2

To remove the radical on the left side of the equation, square both sides of the equation.

${\sqrt{a^{2} + b^{2}}}^{2} = c^{2}$

Step 3

Simplify each side of the equation.

Tap for more steps...

Step 3.1

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

${({(a^{2} + b^{2})}^{\frac{1}{2}})}^{2} = c^{2}$

Step 3.2

Simplify the left side.

Tap for more steps...

Step 3.2.1

Simplify ${({(a^{2} + b^{2})}^{\frac{1}{2}})}^{2}$ .

Tap for more steps...

Step 3.2.1.1

Multiply the exponents in ${({(a^{2} + b^{2})}^{\frac{1}{2}})}^{2}$ .

Tap for more steps...

Step 3.2.1.1.1

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

${(a^{2} + b^{2})}^{\frac{1}{2} \cdot 2} = c^{2}$

Step 3.2.1.1.2

Cancel the common factor of $2$ .

Tap for more steps...

Step 3.2.1.1.2.1

Cancel the common factor.

${(a^{2} + b^{2})}^{\frac{1}{2} \cdot 2} = c^{2}$

Step 3.2.1.1.2.2

Rewrite the expression.

${(a^{2} + b^{2})}^{1} = c^{2}$

Step 3.2.1.2

Simplify.

$a^{2} + b^{2} = c^{2}$

Step 4

Solve for $a$ .

Tap for more steps...

Step 4.1

Subtract $b^{2}$ from both sides of the equation.

$a^{2} = c^{2} - b^{2}$

Step 4.2

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

$a = \pm \sqrt{c^{2} - b^{2}}$

Step 4.3

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

$a = \pm \sqrt{(c + b) (c - b)}$

Step 4.4

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

Tap for more steps...

Step 4.4.1

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

$a = \sqrt{(c + b) (c - b)}$

Step 4.4.2

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

$a = - \sqrt{(c + b) (c - b)}$

Step 4.4.3

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

$a = \sqrt{(c + b) (c - b)}$

$a = - \sqrt{(c + b) (c - b)}$

$a = \sqrt{(c + b) (c - b)}$

$a = - \sqrt{(c + b) (c - b)}$

$a = \sqrt{(c + b) (c - b)}$

$a = - \sqrt{(c + b) (c - b)}$