Precalculus Examples

$x (9 - x) = 20$

Step 1

Move all terms to the left side of the equation and simplify.

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Step 1.1

Simplify the left side.

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Step 1.1.1

Simplify $x (9 - x)$ .

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Step 1.1.1.1

Simplify by multiplying through.

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Step 1.1.1.1.1

Apply the distributive property.

$x \cdot 9 + x (- x) = 20$

Step 1.1.1.1.2

Reorder.

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Step 1.1.1.1.2.1

Move $9$ to the left of $x$ .

$9 \cdot x + x (- x) = 20$

Step 1.1.1.1.2.2

Rewrite using the commutative property of multiplication.

$9 \cdot x - x \cdot x = 20$

Step 1.1.1.2

Multiply $x$ by $x$ by adding the exponents.

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Step 1.1.1.2.1

Move $x$ .

$9 x - (x \cdot x) = 20$

Step 1.1.1.2.2

Multiply $x$ by $x$ .

$9 x - x^{2} = 20$

Step 1.2

Subtract $20$ from both sides of the equation.

$9 x - x^{2} - 20 = 0$

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 $a = - 1$ , $b = 9$ , and $c = - 20$ into the quadratic formula and solve for $x$ .

$\frac{- 9 \pm \sqrt{9^{2} - 4 \cdot (- 1 \cdot - 20)}}{2 \cdot - 1}$

Step 4

Simplify.

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Step 4.1

Simplify the numerator.

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Step 4.1.1

Raise $9$ to the power of $2$ .

$x = \frac{- 9 \pm \sqrt{81 - 4 \cdot - 1 \cdot - 20}}{2 \cdot - 1}$

Step 4.1.2

Multiply $- 4 \cdot - 1 \cdot - 20$ .

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Step 4.1.2.1

Multiply $- 4$ by $- 1$ .

$x = \frac{- 9 \pm \sqrt{81 + 4 \cdot - 20}}{2 \cdot - 1}$

Step 4.1.2.2

Multiply $4$ by $- 20$ .

$x = \frac{- 9 \pm \sqrt{81 - 80}}{2 \cdot - 1}$

Step 4.1.3

Subtract $80$ from $81$ .

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

Step 4.1.4

Any root of $1$ is $1$ .

$x = \frac{- 9 \pm 1}{2 \cdot - 1}$

Step 4.2

Multiply $2$ by $- 1$ .

$x = \frac{- 9 \pm 1}{- 2}$

Step 4.3

Simplify $\frac{- 9 \pm 1}{- 2}$ .

$x = \frac{9 \pm 1}{2}$

Step 5

The final answer is the combination of both solutions.

$x = 5, 4$