Precalculus Examples

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

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 $a = 9$ , $b = - 12$ , and $c = 4$ into the quadratic formula and solve for $x$ .

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

Step 3

Simplify.

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Step 3.1

Simplify the numerator.

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Step 3.1.1

Raise $- 12$ to the power of $2$ .

$x = \frac{12 \pm \sqrt{144 - 4 \cdot 9 \cdot 4}}{2 \cdot 9}$

Step 3.1.2

Multiply $- 4 \cdot 9 \cdot 4$ .

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Step 3.1.2.1

Multiply $- 4$ by $9$ .

$x = \frac{12 \pm \sqrt{144 - 36 \cdot 4}}{2 \cdot 9}$

Step 3.1.2.2

Multiply $- 36$ by $4$ .

$x = \frac{12 \pm \sqrt{144 - 144}}{2 \cdot 9}$

Step 3.1.3

Subtract $144$ from $144$ .

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

Step 3.1.4

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

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

Step 3.1.5

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

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

Step 3.1.6

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

$x = \frac{12}{2 \cdot 9}$

Step 3.2

Multiply $2$ by $9$ .

$x = \frac{12}{18}$

Step 3.3

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

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Step 3.3.1

Factor $6$ out of $12$ .

$x = \frac{6 (2)}{18}$

Step 3.3.2

Cancel the common factors.

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Step 3.3.2.1

Factor $6$ out of $18$ .

$x = \frac{6 \cdot 2}{6 \cdot 3}$

Step 3.3.2.2

Cancel the common factor.

$x = \frac{6 \cdot 2}{6 \cdot 3}$

Step 3.3.2.3

Rewrite the expression.

$x = \frac{2}{3}$

Step 4

The final answer is the combination of both solutions.

$x = \frac{2}{3}$ Double roots