Determine if a Solution y

$y < x^{2} - 2 x - 8$

Insert the values of $x =$ and $y =$ into the equation to find if the ordered pair is a solution.

$y < x^{2} - 2 x - 8$

Rewrite so $x$ is on the left side of the inequality.

$x^{2} - 2 x - 8 > y$

Move $y$ to the left side of the equation by subtracting it from both sides.

$x^{2} - 2 x - 8 - y > 0$

Convert the inequality to an equation.

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

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Substitute the values $a = 1$ , $b = - 2$ , and $c = - 8 - y$ into the quadratic formula and solve for $x$ .

$\frac{2 \pm \sqrt{{(- 2)}^{2} - 4 \cdot (1 \cdot (- 8 - y))}}{2 \cdot 1}$

Simplify.

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Simplify the numerator.

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

$x = \frac{2 \pm \sqrt{4 - 4 \cdot (1 \cdot (- 8 - y))}}{2 \cdot 1}$

Multiply $- 8 - y$ by $1$ .

$x = \frac{2 \pm \sqrt{4 - 4 \cdot (- 8 - y)}}{2 \cdot 1}$

Apply the distributive property.

$x = \frac{2 \pm \sqrt{4 - 4 \cdot - 8 - 4 (- y)}}{2 \cdot 1}$

Multiply $- 4$ by $- 8$ .

$x = \frac{2 \pm \sqrt{4 + 32 - 4 (- y)}}{2 \cdot 1}$

Multiply $- 1$ by $- 4$ .

$x = \frac{2 \pm \sqrt{4 + 32 + 4 y}}{2 \cdot 1}$

Add $4$ and $32$ .

$x = \frac{2 \pm \sqrt{36 + 4 y}}{2 \cdot 1}$

Factor $4$ out of $36 + 4 y$ .

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Factor $4$ out of $36$ .

$x = \frac{2 \pm \sqrt{4 \cdot 9 + 4 y}}{2 \cdot 1}$

Factor $4$ out of $4 \cdot 9 + 4 y$ .

$x = \frac{2 \pm \sqrt{4 (9 + y)}}{2 \cdot 1}$

Rewrite $4 (9 + y)$ as $2^{2} (3^{2} + y)$ .

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

$x = \frac{2 \pm \sqrt{2^{2} (9 + y)}}{2 \cdot 1}$

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

$x = \frac{2 \pm \sqrt{2^{2} (3^{2} + y)}}{2 \cdot 1}$

Pull terms out from under the radical.

$x = \frac{2 \pm 2 \sqrt{3^{2} + y}}{2 \cdot 1}$

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

$x = \frac{2 \pm 2 \sqrt{9 + y}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{2 \pm 2 \sqrt{9 + y}}{2}$

Simplify $\frac{2 \pm 2 \sqrt{9 + y}}{2}$ .

$x = 1 \pm \sqrt{9 + y}$

Simplify the expression to solve for the $+$ portion of the $\pm$ .

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Simplify the numerator.

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

$x = \frac{2 \pm \sqrt{4 - 4 \cdot (1 \cdot (- 8 - y))}}{2 \cdot 1}$

Multiply $- 8 - y$ by $1$ .

$x = \frac{2 \pm \sqrt{4 - 4 \cdot (- 8 - y)}}{2 \cdot 1}$

Apply the distributive property.

$x = \frac{2 \pm \sqrt{4 - 4 \cdot - 8 - 4 (- y)}}{2 \cdot 1}$

Multiply $- 4$ by $- 8$ .

$x = \frac{2 \pm \sqrt{4 + 32 - 4 (- y)}}{2 \cdot 1}$

Multiply $- 1$ by $- 4$ .

$x = \frac{2 \pm \sqrt{4 + 32 + 4 y}}{2 \cdot 1}$

Add $4$ and $32$ .

$x = \frac{2 \pm \sqrt{36 + 4 y}}{2 \cdot 1}$

Factor $4$ out of $36 + 4 y$ .

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Factor $4$ out of $36$ .

$x = \frac{2 \pm \sqrt{4 \cdot 9 + 4 y}}{2 \cdot 1}$

Factor $4$ out of $4 \cdot 9 + 4 y$ .

$x = \frac{2 \pm \sqrt{4 (9 + y)}}{2 \cdot 1}$

Rewrite $4 (9 + y)$ as $2^{2} (3^{2} + y)$ .

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

$x = \frac{2 \pm \sqrt{2^{2} (9 + y)}}{2 \cdot 1}$

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

$x = \frac{2 \pm \sqrt{2^{2} (3^{2} + y)}}{2 \cdot 1}$

Pull terms out from under the radical.

$x = \frac{2 \pm 2 \sqrt{3^{2} + y}}{2 \cdot 1}$

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

$x = \frac{2 \pm 2 \sqrt{9 + y}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{2 \pm 2 \sqrt{9 + y}}{2}$

Simplify $\frac{2 \pm 2 \sqrt{9 + y}}{2}$ .

$x = 1 \pm \sqrt{9 + y}$

Change the $\pm$ to $+$ .

$x = 1 + \sqrt{9 + y}$

Simplify the expression to solve for the $-$ portion of the $\pm$ .

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Simplify the numerator.

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

$x = \frac{2 \pm \sqrt{4 - 4 \cdot (1 \cdot (- 8 - y))}}{2 \cdot 1}$

Multiply $- 8 - y$ by $1$ .

$x = \frac{2 \pm \sqrt{4 - 4 \cdot (- 8 - y)}}{2 \cdot 1}$

Apply the distributive property.

$x = \frac{2 \pm \sqrt{4 - 4 \cdot - 8 - 4 (- y)}}{2 \cdot 1}$

Multiply $- 4$ by $- 8$ .

$x = \frac{2 \pm \sqrt{4 + 32 - 4 (- y)}}{2 \cdot 1}$

Multiply $- 1$ by $- 4$ .

$x = \frac{2 \pm \sqrt{4 + 32 + 4 y}}{2 \cdot 1}$

Add $4$ and $32$ .

$x = \frac{2 \pm \sqrt{36 + 4 y}}{2 \cdot 1}$

Factor $4$ out of $36 + 4 y$ .

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Factor $4$ out of $36$ .

$x = \frac{2 \pm \sqrt{4 \cdot 9 + 4 y}}{2 \cdot 1}$

Factor $4$ out of $4 \cdot 9 + 4 y$ .

$x = \frac{2 \pm \sqrt{4 (9 + y)}}{2 \cdot 1}$

Rewrite $4 (9 + y)$ as $2^{2} (3^{2} + y)$ .

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

$x = \frac{2 \pm \sqrt{2^{2} (9 + y)}}{2 \cdot 1}$

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

$x = \frac{2 \pm \sqrt{2^{2} (3^{2} + y)}}{2 \cdot 1}$

Pull terms out from under the radical.

$x = \frac{2 \pm 2 \sqrt{3^{2} + y}}{2 \cdot 1}$

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

$x = \frac{2 \pm 2 \sqrt{9 + y}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{2 \pm 2 \sqrt{9 + y}}{2}$

Simplify $\frac{2 \pm 2 \sqrt{9 + y}}{2}$ .

$x = 1 \pm \sqrt{9 + y}$

Change the $\pm$ to $-$ .

$x = 1 - \sqrt{9 + y}$

Consolidate the solutions.

$x = 1 + \sqrt{9 + y}$

$x = 1 - \sqrt{9 + y}$

Since the equation is not true when the values are used, the ordered pair is not a solution.

The ordered pair is not a solution to the equation.

Determine if a Solution y<x^2-2x-8

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