Finite Math Examples

$4 x^{2} - 2 x y + 5 y^{2} = 8$

Step 1

Subtract $8$ from both sides of the equation.

$4 x^{2} - 2 x y + 5 y^{2} - 8 = 0$

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 $a = 4$ , $b = - 2 y$ , and $c = 5 y^{2} - 8$ into the quadratic formula and solve for $x$ .

$\frac{2 y \pm \sqrt{{(- 2 y)}^{2} - 4 \cdot (4 \cdot (5 y^{2} - 8))}}{2 \cdot 4}$

Step 4

Simplify.

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Step 4.1

Simplify the numerator.

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Step 4.1.1

Add parentheses.

$x = \frac{2 y \pm \sqrt{{(- 2 y)}^{2} - 4 \cdot (4 \cdot (5 y^{2} - 8))}}{2 \cdot 4}$

Step 4.1.2

Let $u = 4 \cdot (5 y^{2} - 8)$ . Substitute $u$ for all occurrences of $4 \cdot (5 y^{2} - 8)$ .

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Step 4.1.2.1

Apply the product rule to $- 2 y$ .

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

Step 4.1.2.2

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

$x = \frac{2 y \pm \sqrt{4 y^{2} - 4 u}}{2 \cdot 4}$

Step 4.1.3

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

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Step 4.1.3.1

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

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

Step 4.1.3.2

Factor $4$ out of $- 4 u$ .

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

Step 4.1.3.3

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

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

Step 4.1.4

Replace all occurrences of $u$ with $4 \cdot (5 y^{2} - 8)$ .

$x = \frac{2 y \pm \sqrt{4 (y^{2} - (4 \cdot (5 y^{2} - 8)))}}{2 \cdot 4}$

Step 4.1.5

Simplify.

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Step 4.1.5.1

Simplify each term.

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Step 4.1.5.1.1

Apply the distributive property.

$x = \frac{2 y \pm \sqrt{4 (y^{2} - (4 (5 y^{2}) + 4 \cdot - 8))}}{2 \cdot 4}$

Step 4.1.5.1.2

Multiply $5$ by $4$ .

$x = \frac{2 y \pm \sqrt{4 (y^{2} - (20 y^{2} + 4 \cdot - 8))}}{2 \cdot 4}$

Step 4.1.5.1.3

Multiply $4$ by $- 8$ .

$x = \frac{2 y \pm \sqrt{4 (y^{2} - (20 y^{2} - 32))}}{2 \cdot 4}$

Step 4.1.5.1.4

Apply the distributive property.

$x = \frac{2 y \pm \sqrt{4 (y^{2} - (20 y^{2}) + 32)}}{2 \cdot 4}$

Step 4.1.5.1.5

Multiply $20$ by $- 1$ .

$x = \frac{2 y \pm \sqrt{4 (y^{2} - 20 y^{2} + 32)}}{2 \cdot 4}$

Step 4.1.5.1.6

Multiply $- 1$ by $- 32$ .

$x = \frac{2 y \pm \sqrt{4 (y^{2} - 20 y^{2} + 32)}}{2 \cdot 4}$

Step 4.1.5.2

Subtract $20 y^{2}$ from $y^{2}$ .

$x = \frac{2 y \pm \sqrt{4 (- 19 y^{2} + 32)}}{2 \cdot 4}$

Step 4.1.6

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

$x = \frac{2 y \pm \sqrt{2^{2} (- 19 y^{2} + 32)}}{2 \cdot 4}$

Step 4.1.7

Pull terms out from under the radical.

$x = \frac{2 y \pm 2 \sqrt{- 19 y^{2} + 32}}{2 \cdot 4}$

Step 4.2

Multiply $2$ by $4$ .

$x = \frac{2 y \pm 2 \sqrt{- 19 y^{2} + 32}}{8}$

Step 4.3

Simplify $\frac{2 y \pm 2 \sqrt{- 19 y^{2} + 32}}{8}$ .

$x = \frac{y \pm \sqrt{- 19 y^{2} + 32}}{4}$

Step 5

The final answer is the combination of both solutions.

$x = \frac{y + \sqrt{- 19 y^{2} + 32}}{4}$

$x = \frac{y - \sqrt{- 19 y^{2} + 32}}{4}$