Basic Math Examples

$9 z^{2} - 12 z y + 4 y^{2} = 0$

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 $a = 9$ , $b = - 12 y$ , and $c = 4 y^{2}$ into the quadratic formula and solve for $z$ .

$\frac{12 y \pm \sqrt{{(- 12 y)}^{2} - 4 \cdot (9 \cdot (4 y^{2}))}}{2 \cdot 9}$

Step 3

Simplify.

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Step 3.1

Simplify the numerator.

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Step 3.1.1

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

$z = \frac{12 y \pm \sqrt{{(- 12 y)}^{2} - {(2 \cdot 3 \cdot (2 y))}^{2}}}{2 \cdot 9}$

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 $a = - 12 y$ and $b = 2 \cdot 3 \cdot 2 y$ .

$z = \frac{12 y \pm \sqrt{(- 12 y + 2 \cdot 3 \cdot (2 y)) (- 12 y - (2 \cdot 3 \cdot (2 y)))}}{2 \cdot 9}$

Step 3.1.3

Simplify.

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Step 3.1.3.1

Factor $2 y$ out of $- 12 y + 2 \cdot 3 \cdot 2 y$ .

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Step 3.1.3.1.1

Factor $2 y$ out of $- 12 y$ .

$z = \frac{12 y \pm \sqrt{(2 y (- 6) + 2 \cdot 3 \cdot (2 y)) (- 12 y - (2 \cdot 3 \cdot (2 y)))}}{2 \cdot 9}$

Step 3.1.3.1.2

Factor $2 y$ out of $2 \cdot 3 \cdot 2 y$ .

$z = \frac{12 y \pm \sqrt{(2 y (- 6) + 2 y (3 \cdot 2)) (- 12 y - (2 \cdot 3 \cdot (2 y)))}}{2 \cdot 9}$

Step 3.1.3.1.3

Factor $2 y$ out of $2 y (- 6) + 2 y (3 \cdot 2)$ .

$z = \frac{12 y \pm \sqrt{2 y (- 6 + 3 \cdot 2) (- 12 y - (2 \cdot 3 \cdot (2 y)))}}{2 \cdot 9}$

Step 3.1.3.2

Multiply $3$ by $2$ .

$z = \frac{12 y \pm \sqrt{2 y (- 6 + 6) (- 12 y - (2 \cdot 3 \cdot (2 y)))}}{2 \cdot 9}$

Step 3.1.3.3

Combine exponents.

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Step 3.1.3.3.1

Multiply $2$ by $3$ .

$z = \frac{12 y \pm \sqrt{2 y (- 6 + 6) (- 12 y - (6 \cdot (2 y)))}}{2 \cdot 9}$

Step 3.1.3.3.2

Multiply $6$ by $2$ .

$z = \frac{12 y \pm \sqrt{2 y (- 6 + 6) (- 12 y - (12 y))}}{2 \cdot 9}$

Step 3.1.3.3.3

Multiply $12$ by $- 1$ .

$z = \frac{12 y \pm \sqrt{2 y (- 6 + 6) (- 12 y - 12 y)}}{2 \cdot 9}$

Step 3.1.4

Add $- 6$ and $6$ .

$z = \frac{12 y \pm \sqrt{2 y \cdot (0 (- 12 y - 12 y))}}{2 \cdot 9}$

Step 3.1.5

Subtract $12 y$ from $- 12 y$ .

$z = \frac{12 y \pm \sqrt{2 y \cdot 0 \cdot (- 24 y)}}{2 \cdot 9}$

Step 3.1.6

Combine exponents.

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Step 3.1.6.1

Multiply $0$ by $2$ .

$z = \frac{12 y \pm \sqrt{0 y \cdot (- 24 y)}}{2 \cdot 9}$

Step 3.1.6.2

Multiply $0$ by $y$ .

$z = \frac{12 y \pm \sqrt{0 \cdot (- 24 y)}}{2 \cdot 9}$

Step 3.1.6.3

Multiply $0$ by $- 24$ .

$z = \frac{12 y \pm \sqrt{0 y}}{2 \cdot 9}$

Step 3.1.6.4

Multiply $0$ by $y$ .

$z = \frac{12 y \pm \sqrt{0}}{2 \cdot 9}$

Step 3.1.7

Rewrite $0$ as $0^{2}$ .

$z = \frac{12 y \pm \sqrt{0^{2}}}{2 \cdot 9}$

Step 3.1.8

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

$z = \frac{12 y \pm 0}{2 \cdot 9}$

Step 3.1.9

$12 y$ plus or minus $0$ is $12 y$ .

$z = \frac{12 y}{2 \cdot 9}$

Step 3.2

Multiply $2$ by $9$ .

$z = \frac{12 y}{18}$

Step 3.3

Cancel the common factor of $12$ and $18$ .

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Step 3.3.1

Factor $6$ out of $12 y$ .

$z = \frac{6 (2 y)}{18}$

Step 3.3.2

Cancel the common factors.

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Step 3.3.2.1

Factor $6$ out of $18$ .

$z = \frac{6 (2 y)}{6 (3)}$

Step 3.3.2.2

Cancel the common factor.

$z = \frac{6 (2 y)}{6 \cdot 3}$

Step 3.3.2.3

Rewrite the expression.

$z = \frac{2 y}{3}$

Step 4

The final answer is the combination of both solutions.

$z = \frac{2 y}{3}$ Double roots