Precalculus Examples

$x^{2} + y^{2} = 8$ , $y^{2} = 2 x$

Step 1

Solve for $y$ in $y^{2} = 2 x$ .

Tap for more steps...

Step 1.1

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

$y = \pm \sqrt{2 x}$

$x^{2} + y^{2} = 8$

Step 1.2

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

Tap for more steps...

Step 1.2.1

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

$y = \sqrt{2 x}$

$x^{2} + y^{2} = 8$

Step 1.2.2

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

$y = - \sqrt{2 x}$

$x^{2} + y^{2} = 8$

Step 1.2.3

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

$y = \sqrt{2 x}$

$y = - \sqrt{2 x}$

$x^{2} + y^{2} = 8$

$y = \sqrt{2 x}$

$y = - \sqrt{2 x}$

$x^{2} + y^{2} = 8$

$y = \sqrt{2 x}$

$y = - \sqrt{2 x}$

$x^{2} + y^{2} = 8$

Step 2

Solve the system $y = \sqrt{2 x}, x^{2} + y^{2} = 8$ .

Tap for more steps...

Step 2.1

Replace all occurrences of $y$ with $\sqrt{2 x}$ in each equation.

Tap for more steps...

Step 2.1.1

Replace all occurrences of $y$ in $x^{2} + y^{2} = 8$ with $\sqrt{2 x}$ .

$x^{2} + {(\sqrt{2 x})}^{2} = 8$

$y = \sqrt{2 x}$

Step 2.1.2

Simplify the left side.

Tap for more steps...

Step 2.1.2.1

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

Tap for more steps...

Step 2.1.2.1.1

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

$x^{2} + {({(2 x)}^{\frac{1}{2}})}^{2} = 8$

$y = \sqrt{2 x}$

Step 2.1.2.1.2

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

$x^{2} + {(2 x)}^{\frac{1}{2} \cdot 2} = 8$

$y = \sqrt{2 x}$

Step 2.1.2.1.3

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

$x^{2} + {(2 x)}^{\frac{2}{2}} = 8$

$y = \sqrt{2 x}$

Step 2.1.2.1.4

Cancel the common factor of $2$ .

Tap for more steps...

Step 2.1.2.1.4.1

Cancel the common factor.

$x^{2} + {(2 x)}^{\frac{2}{2}} = 8$

$y = \sqrt{2 x}$

Step 2.1.2.1.4.2

Rewrite the expression.

$x^{2} + 2 x = 8$

$y = \sqrt{2 x}$

$x^{2} + 2 x = 8$

$y = \sqrt{2 x}$

Step 2.1.2.1.5

Simplify.

$x^{2} + 2 x = 8$

$y = \sqrt{2 x}$

$x^{2} + 2 x = 8$

$y = \sqrt{2 x}$

$x^{2} + 2 x = 8$

$y = \sqrt{2 x}$

$x^{2} + 2 x = 8$

$y = \sqrt{2 x}$

Step 2.2

Solve for $x$ in $x^{2} + 2 x = 8$ .

Tap for more steps...

Step 2.2.1

Subtract $8$ from both sides of the equation.

$x^{2} + 2 x - 8 = 0$

$y = \sqrt{2 x}$

Step 2.2.2

Factor $x^{2} + 2 x - 8$ using the AC method.

Tap for more steps...

Step 2.2.2.1

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 8$ and whose sum is $2$ .

$- 2, 4$

$y = \sqrt{2 x}$

Step 2.2.2.2

Write the factored form using these integers.

$(x - 2) (x + 4) = 0$

$y = \sqrt{2 x}$

$(x - 2) (x + 4) = 0$

$y = \sqrt{2 x}$

Step 2.2.3

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

$x - 2 = 0$

$x + 4 = 0$

$y = \sqrt{2 x}$

Step 2.2.4

Set $x - 2$ equal to $0$ and solve for $x$ .

Tap for more steps...

Step 2.2.4.1

Set $x - 2$ equal to $0$ .

$x - 2 = 0$

$y = \sqrt{2 x}$

Step 2.2.4.2

Add $2$ to both sides of the equation.

$x = 2$

$y = \sqrt{2 x}$

$x = 2$

$y = \sqrt{2 x}$

Step 2.2.5

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

Tap for more steps...

Step 2.2.5.1

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

$x + 4 = 0$

$y = \sqrt{2 x}$

Step 2.2.5.2

Subtract $4$ from both sides of the equation.

$x = - 4$

$y = \sqrt{2 x}$

$x = - 4$

$y = \sqrt{2 x}$

Step 2.2.6

The final solution is all the values that make $(x - 2) (x + 4) = 0$ true.

$x = 2, - 4$

$y = \sqrt{2 x}$

$x = 2, - 4$

$y = \sqrt{2 x}$

Step 2.3

Replace all occurrences of $x$ with $2$ in each equation.

Tap for more steps...

Step 2.3.1

Replace all occurrences of $x$ in $y = \sqrt{2 x}$ with $2$ .

$y = \sqrt{2 (2)}$

$x = 2$

Step 2.3.2

Simplify the right side.

Tap for more steps...

Step 2.3.2.1

Simplify $\sqrt{2 (2)}$ .

Tap for more steps...

Step 2.3.2.1.1

Multiply $2$ by $2$ .

$y = \sqrt{4}$

$x = 2$

Step 2.3.2.1.2

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

$y = \sqrt{2^{2}}$

$x = 2$

Step 2.3.2.1.3

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

$y = 2$

$x = 2$

$y = 2$

$x = 2$

$y = 2$

$x = 2$

$y = 2$

$x = 2$

Step 2.4

Replace all occurrences of $x$ with $- 4$ in each equation.

Tap for more steps...

Step 2.4.1

Replace all occurrences of $x$ in $y = \sqrt{2 x}$ with $- 4$ .

$y = \sqrt{2 (- 4)}$

$x = - 4$

Step 2.4.2

Simplify the right side.

Tap for more steps...

Step 2.4.2.1

Simplify $\sqrt{2 (- 4)}$ .

Tap for more steps...

Step 2.4.2.1.1

Multiply $2$ by $- 4$ .

$y = \sqrt{- 8}$

$x = - 4$

Step 2.4.2.1.2

Rewrite $- 8$ as $- 1 (8)$ .

$y = \sqrt{- 1 \cdot 8}$

$x = - 4$

Step 2.4.2.1.3

Rewrite $\sqrt{- 1 (8)}$ as $\sqrt{- 1} \cdot \sqrt{8}$ .

$y = \sqrt{- 1} \cdot \sqrt{8}$

$x = - 4$

Step 2.4.2.1.4

Rewrite $\sqrt{- 1}$ as $i$ .

$y = i \cdot \sqrt{8}$

$x = - 4$

Step 2.4.2.1.5

Rewrite $8$ as $2^{2} \cdot 2$ .

Tap for more steps...

Step 2.4.2.1.5.1

Factor $4$ out of $8$ .

$y = i \cdot \sqrt{4 (2)}$

$x = - 4$

Step 2.4.2.1.5.2

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

$y = i \cdot \sqrt{2^{2} \cdot 2}$

$x = - 4$

$y = i \cdot \sqrt{2^{2} \cdot 2}$

$x = - 4$

Step 2.4.2.1.6

Pull terms out from under the radical.

$y = i \cdot (2 \sqrt{2})$

$x = - 4$

Step 2.4.2.1.7

Move $2$ to the left of $i$ .

$y = 2 i \sqrt{2}$

$x = - 4$

$y = 2 i \sqrt{2}$

$x = - 4$

$y = 2 i \sqrt{2}$

$x = - 4$

$y = 2 i \sqrt{2}$

$x = - 4$

$y = 2 i \sqrt{2}$

$x = - 4$

Step 3

Solve the system $y = - \sqrt{2 x}, x^{2} + y^{2} = 8$ .

Tap for more steps...

Step 3.1

Replace all occurrences of $y$ with $- \sqrt{2 x}$ in each equation.

Tap for more steps...

Step 3.1.1

Replace all occurrences of $y$ in $x^{2} + y^{2} = 8$ with $- \sqrt{2 x}$ .

$x^{2} + {(- \sqrt{2 x})}^{2} = 8$

$y = - \sqrt{2 x}$

Step 3.1.2

Simplify the left side.

Tap for more steps...

Step 3.1.2.1

Simplify each term.

Tap for more steps...

Step 3.1.2.1.1

Apply the product rule to $- \sqrt{2 x}$ .

$x^{2} + {(- 1)}^{2} {\sqrt{2 x}}^{2} = 8$

$y = - \sqrt{2 x}$

Step 3.1.2.1.2

Raise $- 1$ to the power of $2$ .

$x^{2} + 1 {\sqrt{2 x}}^{2} = 8$

$y = - \sqrt{2 x}$

Step 3.1.2.1.3

Multiply ${\sqrt{2 x}}^{2}$ by $1$ .

$x^{2} + {\sqrt{2 x}}^{2} = 8$

$y = - \sqrt{2 x}$

Step 3.1.2.1.4

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

Tap for more steps...

Step 3.1.2.1.4.1

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

$x^{2} + {({(2 x)}^{\frac{1}{2}})}^{2} = 8$

$y = - \sqrt{2 x}$

Step 3.1.2.1.4.2

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

$x^{2} + {(2 x)}^{\frac{1}{2} \cdot 2} = 8$

$y = - \sqrt{2 x}$

Step 3.1.2.1.4.3

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

$x^{2} + {(2 x)}^{\frac{2}{2}} = 8$

$y = - \sqrt{2 x}$

Step 3.1.2.1.4.4

Cancel the common factor of $2$ .

Tap for more steps...

Step 3.1.2.1.4.4.1

Cancel the common factor.

$x^{2} + {(2 x)}^{\frac{2}{2}} = 8$

$y = - \sqrt{2 x}$

Step 3.1.2.1.4.4.2

Rewrite the expression.

$x^{2} + 2 x = 8$

$y = - \sqrt{2 x}$

$x^{2} + 2 x = 8$

$y = - \sqrt{2 x}$

Step 3.1.2.1.4.5

Simplify.

$x^{2} + 2 x = 8$

$y = - \sqrt{2 x}$

$x^{2} + 2 x = 8$

$y = - \sqrt{2 x}$

$x^{2} + 2 x = 8$

$y = - \sqrt{2 x}$

$x^{2} + 2 x = 8$

$y = - \sqrt{2 x}$

$x^{2} + 2 x = 8$

$y = - \sqrt{2 x}$

Step 3.2

Solve for $x$ in $x^{2} + 2 x = 8$ .

Tap for more steps...

Step 3.2.1

Subtract $8$ from both sides of the equation.

$x^{2} + 2 x - 8 = 0$

$y = - \sqrt{2 x}$

Step 3.2.2

Factor $x^{2} + 2 x - 8$ using the AC method.

Tap for more steps...

Step 3.2.2.1

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 8$ and whose sum is $2$ .

$- 2, 4$

$y = - \sqrt{2 x}$

Step 3.2.2.2

Write the factored form using these integers.

$(x - 2) (x + 4) = 0$

$y = - \sqrt{2 x}$

$(x - 2) (x + 4) = 0$

$y = - \sqrt{2 x}$

Step 3.2.3

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

$x - 2 = 0$

$x + 4 = 0$

$y = - \sqrt{2 x}$

Step 3.2.4

Set $x - 2$ equal to $0$ and solve for $x$ .

Tap for more steps...

Step 3.2.4.1

Set $x - 2$ equal to $0$ .

$x - 2 = 0$

$y = - \sqrt{2 x}$

Step 3.2.4.2

Add $2$ to both sides of the equation.

$x = 2$

$y = - \sqrt{2 x}$

$x = 2$

$y = - \sqrt{2 x}$

Step 3.2.5

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

Tap for more steps...

Step 3.2.5.1

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

$x + 4 = 0$

$y = - \sqrt{2 x}$

Step 3.2.5.2

Subtract $4$ from both sides of the equation.

$x = - 4$

$y = - \sqrt{2 x}$

$x = - 4$

$y = - \sqrt{2 x}$

Step 3.2.6

The final solution is all the values that make $(x - 2) (x + 4) = 0$ true.

$x = 2, - 4$

$y = - \sqrt{2 x}$

$x = 2, - 4$

$y = - \sqrt{2 x}$

Step 3.3

Replace all occurrences of $x$ with $2$ in each equation.

Tap for more steps...

Step 3.3.1

Replace all occurrences of $x$ in $y = - \sqrt{2 x}$ with $2$ .

$y = - \sqrt{2 (2)}$

$x = 2$

Step 3.3.2

Simplify the right side.

Tap for more steps...

Step 3.3.2.1

Simplify $- \sqrt{2 (2)}$ .

Tap for more steps...

Step 3.3.2.1.1

Multiply $2$ by $2$ .

$y = - \sqrt{4}$

$x = 2$

Step 3.3.2.1.2

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

$y = - \sqrt{2^{2}}$

$x = 2$

Step 3.3.2.1.3

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

$y = - 1 \cdot 2$

$x = 2$

Step 3.3.2.1.4

Multiply $- 1$ by $2$ .

$y = - 2$

$x = 2$

$y = - 2$

$x = 2$

$y = - 2$

$x = 2$

$y = - 2$

$x = 2$

Step 3.4

Replace all occurrences of $x$ with $- 4$ in each equation.

Tap for more steps...

Step 3.4.1

Replace all occurrences of $x$ in $y = - \sqrt{2 x}$ with $- 4$ .

$y = - \sqrt{2 (- 4)}$

$x = - 4$

Step 3.4.2

Simplify the right side.

Tap for more steps...

Step 3.4.2.1

Simplify $- \sqrt{2 (- 4)}$ .

Tap for more steps...

Step 3.4.2.1.1

Multiply $2$ by $- 4$ .

$y = - \sqrt{- 8}$

$x = - 4$

Step 3.4.2.1.2

Rewrite $- 8$ as $- 1 (8)$ .

$y = - \sqrt{- 1 \cdot 8}$

$x = - 4$

Step 3.4.2.1.3

Rewrite $\sqrt{- 1 (8)}$ as $\sqrt{- 1} \cdot \sqrt{8}$ .

$y = - (\sqrt{- 1} \cdot \sqrt{8})$

$x = - 4$

Step 3.4.2.1.4

Rewrite $\sqrt{- 1}$ as $i$ .

$y = - (i \cdot \sqrt{8})$

$x = - 4$

Step 3.4.2.1.5

Rewrite $8$ as $2^{2} \cdot 2$ .

Tap for more steps...

Step 3.4.2.1.5.1

Factor $4$ out of $8$ .

$y = - (i \cdot \sqrt{4 (2)})$

$x = - 4$

Step 3.4.2.1.5.2

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

$y = - (i \cdot \sqrt{2^{2} \cdot 2})$

$x = - 4$

$y = - (i \cdot \sqrt{2^{2} \cdot 2})$

$x = - 4$

Step 3.4.2.1.6

Pull terms out from under the radical.

$y = - (i \cdot (2 \sqrt{2}))$

$x = - 4$

Step 3.4.2.1.7

Simplify the expression.

Tap for more steps...

Step 3.4.2.1.7.1

Move $2$ to the left of $i$ .

$y = - (2 \cdot (i \sqrt{2}))$

$x = - 4$

Step 3.4.2.1.7.2

Multiply $2$ by $- 1$ .

$y = - 2 i \sqrt{2}$

$x = - 4$

$y = - 2 i \sqrt{2}$

$x = - 4$

$y = - 2 i \sqrt{2}$

$x = - 4$

$y = - 2 i \sqrt{2}$

$x = - 4$

$y = - 2 i \sqrt{2}$

$x = - 4$

$y = - 2 i \sqrt{2}$

$x = - 4$

Step 4

List all of the solutions.

$y = 2, x = 2$

$y = 2 i \sqrt{2}, x = - 4$

$y = - 2, x = 2$

$y = - 2 i \sqrt{2}, x = - 4$

Step 5