Precalculus Examples

$6 x (x - 1) = 1 - 7 x$

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 $6 x (x - 1)$ .

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Step 1.1.1.1

Apply the distributive property.

$6 x \cdot x + 6 x \cdot - 1 = 1 - 7 x$

Step 1.1.1.2

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

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Step 1.1.1.2.1

Move $x$ .

$6 (x \cdot x) + 6 x \cdot - 1 = 1 - 7 x$

Step 1.1.1.2.2

Multiply $x$ by $x$ .

$6 x^{2} + 6 x \cdot - 1 = 1 - 7 x$

Step 1.1.1.3

Multiply $- 1$ by $6$ .

$6 x^{2} - 6 x = 1 - 7 x$

Step 1.2

Move all the expressions to the left side of the equation.

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Step 1.2.1

Subtract $1$ from both sides of the equation.

$6 x^{2} - 6 x - 1 = - 7 x$

Step 1.2.2

Add $7 x$ to both sides of the equation.

$6 x^{2} - 6 x - 1 + 7 x = 0$

Step 1.3

Add $- 6 x$ and $7 x$ .

$6 x^{2} + x - 1 = 0$

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 $a = 6$ , $b = 1$ , and $c = - 1$ into the quadratic formula and solve for $x$ .

$\frac{- 1 \pm \sqrt{1^{2} - 4 \cdot (6 \cdot - 1)}}{2 \cdot 6}$

Step 4

Simplify.

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Step 4.1

Simplify the numerator.

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Step 4.1.1

One to any power is one.

$x = \frac{- 1 \pm \sqrt{1 - 4 \cdot 6 \cdot - 1}}{2 \cdot 6}$

Step 4.1.2

Multiply $- 4 \cdot 6 \cdot - 1$ .

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Step 4.1.2.1

Multiply $- 4$ by $6$ .

$x = \frac{- 1 \pm \sqrt{1 - 24 \cdot - 1}}{2 \cdot 6}$

Step 4.1.2.2

Multiply $- 24$ by $- 1$ .

$x = \frac{- 1 \pm \sqrt{1 + 24}}{2 \cdot 6}$

Step 4.1.3

Add $1$ and $24$ .

$x = \frac{- 1 \pm \sqrt{25}}{2 \cdot 6}$

Step 4.1.4

Rewrite $25$ as $5^{2}$ .

$x = \frac{- 1 \pm \sqrt{5^{2}}}{2 \cdot 6}$

Step 4.1.5

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

$x = \frac{- 1 \pm 5}{2 \cdot 6}$

Step 4.2

Multiply $2$ by $6$ .

$x = \frac{- 1 \pm 5}{12}$

Step 5

The final answer is the combination of both solutions.

$x = \frac{1}{3}, - \frac{1}{2}$