Algebra Examples

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

Step 1

Use the quadratic formula to find the solutions.

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

Step 2

Substitute the values $a = 1$ , $b = - 5 y$ , and $c = 4 y^{2}$ into the quadratic formula and solve for $x$ .

$\frac{5 y \pm \sqrt{{(- 5 y)}^{2} - 4 \cdot (1 \cdot (4 y^{2}))}}{2 \cdot 1}$

Step 3

Simplify.

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Step 3.1

Simplify the numerator.

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Step 3.1.1

Rewrite $4 \cdot 1 \cdot 4 y^{2}$ as ${(2 \cdot 1 \cdot 2 y)}^{2}$ .

$x = \frac{5 y \pm \sqrt{{(- 5 y)}^{2} - {(2 \cdot 1 \cdot (2 y))}^{2}}}{2 \cdot 1}$

Step 3.1.2

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = - 5 y$ and $b = 2 \cdot 1 \cdot 2 y$ .

$x = \frac{5 y \pm \sqrt{(- 5 y + 2 \cdot 1 \cdot (2 y)) (- 5 y - (2 \cdot 1 \cdot (2 y)))}}{2 \cdot 1}$

Step 3.1.3

Simplify.

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Step 3.1.3.1

Factor $y$ out of $- 5 y + 2 \cdot 1 \cdot 2 y$ .

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Step 3.1.3.1.1

Factor $y$ out of $- 5 y$ .

$x = \frac{5 y \pm \sqrt{(y \cdot - 5 + 2 \cdot 1 \cdot (2 y)) (- 5 y - (2 \cdot 1 \cdot (2 y)))}}{2 \cdot 1}$

Step 3.1.3.1.2

Factor $y$ out of $2 \cdot 1 \cdot 2 y$ .

$x = \frac{5 y \pm \sqrt{(y \cdot - 5 + y (2 \cdot 1 \cdot 2)) (- 5 y - (2 \cdot 1 \cdot (2 y)))}}{2 \cdot 1}$

Step 3.1.3.1.3

Factor $y$ out of $y \cdot - 5 + y (2 \cdot 1 \cdot 2)$ .

$x = \frac{5 y \pm \sqrt{y (- 5 + 2 \cdot 1 \cdot 2) (- 5 y - (2 \cdot 1 \cdot (2 y)))}}{2 \cdot 1}$

Step 3.1.3.2

Multiply $2 \cdot 1 \cdot 2$ .

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Step 3.1.3.2.1

Multiply $2$ by $1$ .

$x = \frac{5 y \pm \sqrt{y (- 5 + 2 \cdot 2) (- 5 y - (2 \cdot 1 \cdot (2 y)))}}{2 \cdot 1}$

Step 3.1.3.2.2

Multiply $2$ by $2$ .

$x = \frac{5 y \pm \sqrt{y (- 5 + 4) (- 5 y - (2 \cdot 1 \cdot (2 y)))}}{2 \cdot 1}$

Step 3.1.3.3

Add $- 5$ and $4$ .

$x = \frac{5 y \pm \sqrt{y \cdot (- 1 (- 5 y - (2 \cdot 1 \cdot (2 y))))}}{2 \cdot 1}$

Step 3.1.3.4

Combine exponents.

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Step 3.1.3.4.1

Factor out negative.

$x = \frac{5 y \pm \sqrt{- (y (- 5 y - (2 \cdot 1 \cdot (2 y))))}}{2 \cdot 1}$

Step 3.1.3.4.2

Multiply $2$ by $1$ .

$x = \frac{5 y \pm \sqrt{- (y (- 5 y - (2 \cdot (2 y))))}}{2 \cdot 1}$

Step 3.1.3.4.3

Multiply $2$ by $2$ .

$x = \frac{5 y \pm \sqrt{- (y (- 5 y - (4 y)))}}{2 \cdot 1}$

Step 3.1.3.4.4

Multiply $4$ by $- 1$ .

$x = \frac{5 y \pm \sqrt{- (y (- 5 y - 4 y))}}{2 \cdot 1}$

$x = \frac{5 y \pm \sqrt{- y (- 5 y - 4 y)}}{2 \cdot 1}$

Step 3.1.4

Subtract $4 y$ from $- 5 y$ .

$x = \frac{5 y \pm \sqrt{- y \cdot (- 9 y)}}{2 \cdot 1}$

Step 3.1.5

Combine exponents.

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Step 3.1.5.1

Multiply $- 9$ by $- 1$ .

$x = \frac{5 y \pm \sqrt{9 y \cdot y}}{2 \cdot 1}$

Step 3.1.5.2

Raise $y$ to the power of $1$ .

$x = \frac{5 y \pm \sqrt{9 (y \cdot y)}}{2 \cdot 1}$

Step 3.1.5.3

Raise $y$ to the power of $1$ .

$x = \frac{5 y \pm \sqrt{9 (y \cdot y)}}{2 \cdot 1}$

Step 3.1.5.4

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$x = \frac{5 y \pm \sqrt{9 y^{1 + 1}}}{2 \cdot 1}$

Step 3.1.5.5

Add $1$ and $1$ .

$x = \frac{5 y \pm \sqrt{9 y^{2}}}{2 \cdot 1}$

Step 3.1.6

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

$x = \frac{5 y \pm \sqrt{{(3 y)}^{2}}}{2 \cdot 1}$

Step 3.1.7

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

$x = \frac{5 y \pm 3 y}{2 \cdot 1}$

Step 3.2

Multiply $2$ by $1$ .

$x = \frac{5 y \pm 3 y}{2}$

Step 4

The final answer is the combination of both solutions.

$x = 4 y$

$x = y$