Precalculus Examples

$\sqrt{3 z^{2} + 16} = \sqrt{2 z^{2} - 8 z}$

Step 1

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

${\sqrt{3 z^{2} + 16}}^{2} = {\sqrt{2 z^{2} - 8 z}}^{2}$

Step 2

Simplify each side of the equation.

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Step 2.1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{3 z^{2} + 16}$ as ${(3 z^{2} + 16)}^{\frac{1}{2}}$ .

${({(3 z^{2} + 16)}^{\frac{1}{2}})}^{2} = {\sqrt{2 z^{2} - 8 z}}^{2}$

Step 2.2

Simplify the left side.

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Step 2.2.1

Simplify ${({(3 z^{2} + 16)}^{\frac{1}{2}})}^{2}$ .

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Step 2.2.1.1

Multiply the exponents in ${({(3 z^{2} + 16)}^{\frac{1}{2}})}^{2}$ .

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Step 2.2.1.1.1

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

${(3 z^{2} + 16)}^{\frac{1}{2} \cdot 2} = {\sqrt{2 z^{2} - 8 z}}^{2}$

Step 2.2.1.1.2

Cancel the common factor of $2$ .

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Step 2.2.1.1.2.1

Cancel the common factor.

${(3 z^{2} + 16)}^{\frac{1}{2} \cdot 2} = {\sqrt{2 z^{2} - 8 z}}^{2}$

Step 2.2.1.1.2.2

Rewrite the expression.

${(3 z^{2} + 16)}^{1} = {\sqrt{2 z^{2} - 8 z}}^{2}$

Step 2.2.1.2

Simplify.

$3 z^{2} + 16 = {\sqrt{2 z^{2} - 8 z}}^{2}$

Step 2.3

Simplify the right side.

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Step 2.3.1

Simplify ${\sqrt{2 z^{2} - 8 z}}^{2}$ .

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Step 2.3.1.1

Factor $2 z$ out of $2 z^{2} - 8 z$ .

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Step 2.3.1.1.1

Factor $2 z$ out of $2 z^{2}$ .

$3 z^{2} + 16 = {\sqrt{2 z (z) - 8 z}}^{2}$

Step 2.3.1.1.2

Factor $2 z$ out of $- 8 z$ .

$3 z^{2} + 16 = {\sqrt{2 z (z) + 2 z (- 4)}}^{2}$

Step 2.3.1.1.3

Factor $2 z$ out of $2 z (z) + 2 z (- 4)$ .

$3 z^{2} + 16 = {\sqrt{2 z (z - 4)}}^{2}$

Step 2.3.1.2

Rewrite ${\sqrt{2 z (z - 4)}}^{2}$ as $2 z (z - 4)$ .

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Step 2.3.1.2.1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{2 z (z - 4)}$ as ${(2 z (z - 4))}^{\frac{1}{2}}$ .

$3 z^{2} + 16 = {({(2 z (z - 4))}^{\frac{1}{2}})}^{2}$

Step 2.3.1.2.2

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

$3 z^{2} + 16 = {(2 z (z - 4))}^{\frac{1}{2} \cdot 2}$

Step 2.3.1.2.3

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

$3 z^{2} + 16 = {(2 z (z - 4))}^{\frac{2}{2}}$

Step 2.3.1.2.4

Cancel the common factor of $2$ .

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Step 2.3.1.2.4.1

Cancel the common factor.

$3 z^{2} + 16 = {(2 z (z - 4))}^{\frac{2}{2}}$

Step 2.3.1.2.4.2

Rewrite the expression.

$3 z^{2} + 16 = {(2 z (z - 4))}^{1}$

Step 2.3.1.2.5

Simplify.

$3 z^{2} + 16 = 2 z (z - 4)$

Step 2.3.1.3

Apply the distributive property.

$3 z^{2} + 16 = 2 z \cdot z + 2 z \cdot - 4$

Step 2.3.1.4

Multiply $z$ by $z$ by adding the exponents.

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Step 2.3.1.4.1

Move $z$ .

$3 z^{2} + 16 = 2 (z \cdot z) + 2 z \cdot - 4$

Step 2.3.1.4.2

Multiply $z$ by $z$ .

$3 z^{2} + 16 = 2 z^{2} + 2 z \cdot - 4$

Step 2.3.1.5

Multiply $- 4$ by $2$ .

$3 z^{2} + 16 = 2 z^{2} - 8 z$

Step 3

Solve for $z$ .

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Step 3.1

Since $z$ is on the right side of the equation, switch the sides so it is on the left side of the equation.

$2 z^{2} - 8 z = 3 z^{2} + 16$

Step 3.2

Move all terms containing $z$ to the left side of the equation.

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Step 3.2.1

Subtract $3 z^{2}$ from both sides of the equation.

$2 z^{2} - 8 z - 3 z^{2} = 16$

Step 3.2.2

Subtract $3 z^{2}$ from $2 z^{2}$ .

$- z^{2} - 8 z = 16$

Step 3.3

Subtract $16$ from both sides of the equation.

$- z^{2} - 8 z - 16 = 0$

Step 3.4

Factor the left side of the equation.

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Step 3.4.1

Factor $- 1$ out of $- z^{2} - 8 z - 16$ .

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Step 3.4.1.1

Factor $- 1$ out of $- z^{2}$ .

$- (z^{2}) - 8 z - 16 = 0$

Step 3.4.1.2

Factor $- 1$ out of $- 8 z$ .

$- (z^{2}) - (8 z) - 16 = 0$

Step 3.4.1.3

Rewrite $- 16$ as $- 1 (16)$ .

$- (z^{2}) - (8 z) - 1 \cdot 16 = 0$

Step 3.4.1.4

Factor $- 1$ out of $- (z^{2}) - (8 z)$ .

$- (z^{2} + 8 z) - 1 \cdot 16 = 0$

Step 3.4.1.5

Factor $- 1$ out of $- (z^{2} + 8 z) - 1 (16)$ .

$- (z^{2} + 8 z + 16) = 0$

Step 3.4.2

Factor using the perfect square rule.

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Step 3.4.2.1

Rewrite $16$ as $4^{2}$ .

$- (z^{2} + 8 z + 4^{2}) = 0$

Step 3.4.2.2

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$8 z = 2 \cdot z \cdot 4$

Step 3.4.2.3

Rewrite the polynomial.

$- (z^{2} + 2 \cdot z \cdot 4 + 4^{2}) = 0$

Step 3.4.2.4

Factor using the perfect square trinomial rule $a^{2} + 2 a b + b^{2} = {(a + b)}^{2}$ , where $a = z$ and $b = 4$ .

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

Step 3.5

Divide each term in $- {(z + 4)}^{2} = 0$ by $- 1$ and simplify.

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Step 3.5.1

Divide each term in $- {(z + 4)}^{2} = 0$ by $- 1$ .

$\frac{- {(z + 4)}^{2}}{- 1} = \frac{0}{- 1}$

Step 3.5.2

Simplify the left side.

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Step 3.5.2.1

Dividing two negative values results in a positive value.

$\frac{{(z + 4)}^{2}}{1} = \frac{0}{- 1}$

Step 3.5.2.2

Divide ${(z + 4)}^{2}$ by $1$ .

${(z + 4)}^{2} = \frac{0}{- 1}$

Step 3.5.3

Simplify the right side.

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Step 3.5.3.1

Divide $0$ by $- 1$ .

${(z + 4)}^{2} = 0$

Step 3.6

Set the $z + 4$ equal to $0$ .

$z + 4 = 0$

Step 3.7

Subtract $4$ from both sides of the equation.

$z = - 4$