Finite Math Examples

$x^{2} + y^{2} + 2 x + 4 y = 11$

Step 1

Subtract $11$ from both sides of the equation.

$x^{2} + y^{2} + 2 x + 4 y - 11 = 0$

Step 2

Use the quadratic formula to find the solutions.

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

Step 3

Substitute the values $a = 1$ , $b = 2$ , and $c = y^{2} + 4 y - 11$ into the quadratic formula and solve for $x$ .

$\frac{- 2 \pm \sqrt{2^{2} - 4 \cdot (1 \cdot (y^{2} + 4 y - 11))}}{2 \cdot 1}$

Step 4

Simplify.

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Step 4.1

Simplify the numerator.

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Step 4.1.1

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

$x = \frac{- 2 \pm \sqrt{4 - 4 \cdot 1 \cdot (y^{2} + 4 y - 11)}}{2 \cdot 1}$

Step 4.1.2

Multiply $- 4$ by $1$ .

$x = \frac{- 2 \pm \sqrt{4 - 4 \cdot (y^{2} + 4 y - 11)}}{2 \cdot 1}$

Step 4.1.3

Apply the distributive property.

$x = \frac{- 2 \pm \sqrt{4 - 4 y^{2} - 4 (4 y) - 4 \cdot - 11}}{2 \cdot 1}$

Step 4.1.4

Simplify.

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Step 4.1.4.1

Multiply $4$ by $- 4$ .

$x = \frac{- 2 \pm \sqrt{4 - 4 y^{2} - 16 y - 4 \cdot - 11}}{2 \cdot 1}$

Step 4.1.4.2

Multiply $- 4$ by $- 11$ .

$x = \frac{- 2 \pm \sqrt{4 - 4 y^{2} - 16 y + 44}}{2 \cdot 1}$

Step 4.1.5

Add $4$ and $44$ .

$x = \frac{- 2 \pm \sqrt{- 4 y^{2} - 16 y + 48}}{2 \cdot 1}$

Step 4.1.6

Rewrite $- 4 y^{2} - 16 y + 48$ in a factored form.

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Step 4.1.6.1

Factor $4$ out of $- 4 y^{2} - 16 y + 48$ .

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Step 4.1.6.1.1

Factor $4$ out of $- 4 y^{2}$ .

$x = \frac{- 2 \pm \sqrt{4 (- y^{2}) - 16 y + 48}}{2 \cdot 1}$

Step 4.1.6.1.2

Factor $4$ out of $- 16 y$ .

$x = \frac{- 2 \pm \sqrt{4 (- y^{2}) + 4 (- 4 y) + 48}}{2 \cdot 1}$

Step 4.1.6.1.3

Factor $4$ out of $48$ .

$x = \frac{- 2 \pm \sqrt{4 (- y^{2}) + 4 (- 4 y) + 4 (12)}}{2 \cdot 1}$

Step 4.1.6.1.4

Factor $4$ out of $4 (- y^{2}) + 4 (- 4 y)$ .

$x = \frac{- 2 \pm \sqrt{4 (- y^{2} - 4 y) + 4 (12)}}{2 \cdot 1}$

Step 4.1.6.1.5

Factor $4$ out of $4 (- y^{2} - 4 y) + 4 (12)$ .

$x = \frac{- 2 \pm \sqrt{4 (- y^{2} - 4 y + 12)}}{2 \cdot 1}$

Step 4.1.6.2

Factor by grouping.

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Step 4.1.6.2.1

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = - 1 \cdot 12 = - 12$ and whose sum is $b = - 4$ .

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Step 4.1.6.2.1.1

Factor $- 4$ out of $- 4 y$ .

$x = \frac{- 2 \pm \sqrt{4 (- y^{2} - 4 y + 12)}}{2 \cdot 1}$

Step 4.1.6.2.1.2

Rewrite $- 4$ as $2$ plus $- 6$

$x = \frac{- 2 \pm \sqrt{4 (- y^{2} + (2 - 6) y + 12)}}{2 \cdot 1}$

Step 4.1.6.2.1.3

Apply the distributive property.

$x = \frac{- 2 \pm \sqrt{4 (- y^{2} + 2 y - 6 y + 12)}}{2 \cdot 1}$

Step 4.1.6.2.2

Factor out the greatest common factor from each group.

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Step 4.1.6.2.2.1

Group the first two terms and the last two terms.

$x = \frac{- 2 \pm \sqrt{4 ((- y^{2} + 2 y) - 6 y + 12)}}{2 \cdot 1}$

Step 4.1.6.2.2.2

Factor out the greatest common factor (GCF) from each group.

$x = \frac{- 2 \pm \sqrt{4 (y (- y + 2) + 6 (- y + 2))}}{2 \cdot 1}$

Step 4.1.6.2.3

Factor the polynomial by factoring out the greatest common factor, $- y + 2$ .

$x = \frac{- 2 \pm \sqrt{4 ((- y + 2) (y + 6))}}{2 \cdot 1}$

$x = \frac{- 2 \pm \sqrt{4 (- y + 2) (y + 6)}}{2 \cdot 1}$

Step 4.1.7

Rewrite $4 (- y + 2) (y + 6)$ as $2^{2} ((- y + 2) (y + 6))$ .

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Step 4.1.7.1

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

$x = \frac{- 2 \pm \sqrt{2^{2} (- y + 2) (y + 6)}}{2 \cdot 1}$

Step 4.1.7.2

Add parentheses.

$x = \frac{- 2 \pm \sqrt{2^{2} ((- y + 2) (y + 6))}}{2 \cdot 1}$

Step 4.1.8

Pull terms out from under the radical.

$x = \frac{- 2 \pm 2 \sqrt{(- y + 2) (y + 6)}}{2 \cdot 1}$

Step 4.2

Multiply $2$ by $1$ .

$x = \frac{- 2 \pm 2 \sqrt{(- y + 2) (y + 6)}}{2}$

Step 4.3

Simplify $\frac{- 2 \pm 2 \sqrt{(- y + 2) (y + 6)}}{2}$ .

$x = - 1 \pm \sqrt{(- y + 2) (y + 6)}$

Step 5

The final answer is the combination of both solutions.

$x = - 1 + \sqrt{(- y + 2) (y + 6)}$

$x = - 1 - \sqrt{(- y + 2) (y + 6)}$