Precalculus Examples

$x^{2} + 3 x - \frac{7}{4} = 0$

Step 1

Add $\frac{7}{4}$ to both sides of the equation.

$x^{2} + 3 x = \frac{7}{4}$

Step 2

To create a trinomial square on the left side of the equation, find a value that is equal to the square of half of $b$ .

${(\frac{b}{2})}^{2} = {(\frac{3}{2})}^{2}$

Step 3

Add the term to each side of the equation.

$x^{2} + 3 x + {(\frac{3}{2})}^{2} = \frac{7}{4} + {(\frac{3}{2})}^{2}$

Step 4

Simplify the equation.

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Step 4.1

Simplify the left side.

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Step 4.1.1

Simplify each term.

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Step 4.1.1.1

Apply the product rule to $\frac{3}{2}$ .

$x^{2} + 3 x + \frac{3^{2}}{2^{2}} = \frac{7}{4} + {(\frac{3}{2})}^{2}$

Step 4.1.1.2

Raise $3$ to the power of $2$ .

$x^{2} + 3 x + \frac{9}{2^{2}} = \frac{7}{4} + {(\frac{3}{2})}^{2}$

Step 4.1.1.3

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

$x^{2} + 3 x + \frac{9}{4} = \frac{7}{4} + {(\frac{3}{2})}^{2}$

Step 4.2

Simplify the right side.

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Step 4.2.1

Simplify $\frac{7}{4} + {(\frac{3}{2})}^{2}$ .

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Step 4.2.1.1

Simplify each term.

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Step 4.2.1.1.1

Apply the product rule to $\frac{3}{2}$ .

$x^{2} + 3 x + \frac{9}{4} = \frac{7}{4} + \frac{3^{2}}{2^{2}}$

Step 4.2.1.1.2

Raise $3$ to the power of $2$ .

$x^{2} + 3 x + \frac{9}{4} = \frac{7}{4} + \frac{9}{2^{2}}$

Step 4.2.1.1.3

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

$x^{2} + 3 x + \frac{9}{4} = \frac{7}{4} + \frac{9}{4}$

Step 4.2.1.2

Combine fractions.

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Step 4.2.1.2.1

Combine the numerators over the common denominator.

$x^{2} + 3 x + \frac{9}{4} = \frac{7 + 9}{4}$

Step 4.2.1.2.2

Simplify the expression.

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Step 4.2.1.2.2.1

Add $7$ and $9$ .

$x^{2} + 3 x + \frac{9}{4} = \frac{16}{4}$

Step 4.2.1.2.2.2

Divide $16$ by $4$ .

$x^{2} + 3 x + \frac{9}{4} = 4$

Step 5

Factor the perfect trinomial square into ${(x + \frac{3}{2})}^{2}$ .

${(x + \frac{3}{2})}^{2} = 4$

Step 6

Solve the equation for $x$ .

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Step 6.1

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

$x + \frac{3}{2} = \pm \sqrt{4}$

Step 6.2

Simplify $\pm \sqrt{4}$ .

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Step 6.2.1

Rewrite $4$ as $2^{2}$ .

$x + \frac{3}{2} = \pm \sqrt{2^{2}}$

Step 6.2.2

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

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

Step 6.3

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

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Step 6.3.1

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

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

Step 6.3.2

Move all terms not containing $x$ to the right side of the equation.

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Step 6.3.2.1

Subtract $\frac{3}{2}$ from both sides of the equation.

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

Step 6.3.2.2

To write $2$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

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

Step 6.3.2.3

Combine $2$ and $\frac{2}{2}$ .

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

Step 6.3.2.4

Combine the numerators over the common denominator.

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

Step 6.3.2.5

Simplify the numerator.

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Step 6.3.2.5.1

Multiply $2$ by $2$ .

$x = \frac{4 - 3}{2}$

Step 6.3.2.5.2

Subtract $3$ from $4$ .

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

Step 6.3.3

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

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

Step 6.3.4

Move all terms not containing $x$ to the right side of the equation.

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Step 6.3.4.1

Subtract $\frac{3}{2}$ from both sides of the equation.

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

Step 6.3.4.2

To write $- 2$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

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

Step 6.3.4.3

Combine $- 2$ and $\frac{2}{2}$ .

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

Step 6.3.4.4

Combine the numerators over the common denominator.

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

Step 6.3.4.5

Simplify the numerator.

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Step 6.3.4.5.1

Multiply $- 2$ by $2$ .

$x = \frac{- 4 - 3}{2}$

Step 6.3.4.5.2

Subtract $3$ from $- 4$ .

$x = \frac{- 7}{2}$

Step 6.3.4.6

Move the negative in front of the fraction.

$x = - \frac{7}{2}$

Step 6.3.5

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

$x = \frac{1}{2}, - \frac{7}{2}$