Precalculus Examples

$| x | = 38$ , $x = 45$

Step 1

Remove the absolute value term. This creates a $\pm$ on the right side of the equation because $| x | = \pm x$ .

$x = \pm 38$

$x = 45$

Step 2

The complete solution is the result of both the positive and negative portions of the solution.

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Step 2.1

First, use the positive value of the $\pm$ to find the first solution.

$x = 38$

$x = 45$

Step 2.2

Next, use the negative value of the $\pm$ to find the second solution.

$x = - 38$

$x = 45$

Step 2.3

The complete solution is the result of both the positive and negative portions of the solution.

$x = 38, - 38$

$x = 45$

$x = 38, - 38$

$x = 45$

Step 3

Since the roots of an equation are the points where the solution is $0$ , set each root as a factor of the equation that equals $0$ .

$(x - (38, - 38)) (x - 45) = 0$

Step 4

Simplify.

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Step 4.1

Simplify by multiplying through.

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Step 4.1.1

Apply the distributive property.

$(x - 38, - 38) x + (x - 38, - 38) \cdot - 45 = 0$

Step 4.1.2

Move $- 45$ to the left of $(x - 38, - 38)$ .

$(x - 38, - 38) x - 45 \cdot (x - 38, - 38) = 0$

Step 4.2

Simplify each term.

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Step 4.2.1

Multiply $- 45$ by each element of the matrix.

$(x - 38, - 38) x + (- 45 (x - 38), - 45 \cdot - 38) = 0$

Step 4.2.2

Simplify each element in the matrix.

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Step 4.2.2.1

Apply the distributive property.

$(x - 38, - 38) x + (- 45 x - 45 \cdot - 38, - 45 \cdot - 38) = 0$

Step 4.2.2.2

Multiply $- 45$ by $- 38$ .

$(x - 38, - 38) x + (- 45 x + 1710, - 45 \cdot - 38) = 0$

Step 4.2.2.3

Multiply $- 45$ by $- 38$ .

$(x - 38, - 38) x + (- 45 x + 1710, 1710) = 0$

Step 4.3

Reorder factors in $(x - 38, - 38) x + (- 45 x + 1710, 1710)$ .

$x (x - 38, - 38) + (- 45 x + 1710, 1710) = 0$