Pre-Algebra Examples

${(x - \frac{3}{4})}^{2} = - \frac{39}{16}$

Step 1

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

$x - \frac{3}{4} = \pm \sqrt{- \frac{39}{16}}$

Step 2

Simplify $\pm \sqrt{- \frac{39}{16}}$ .

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Step 2.1

Rewrite $- \frac{39}{16}$ as ${(\frac{i}{4})}^{2} \cdot 39$ .

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Step 2.1.1

Rewrite $- 1$ as $i^{2}$ .

$x - \frac{3}{4} = \pm \sqrt{i^{2} \frac{39}{16}}$

Step 2.1.2

Factor the perfect power $1^{2}$ out of $39$ .

$x - \frac{3}{4} = \pm \sqrt{i^{2} \frac{1^{2} \cdot 39}{16}}$

Step 2.1.3

Factor the perfect power $4^{2}$ out of $16$ .

$x - \frac{3}{4} = \pm \sqrt{i^{2} \frac{1^{2} \cdot 39}{4^{2} \cdot 1}}$

Step 2.1.4

Rearrange the fraction $\frac{1^{2} \cdot 39}{4^{2} \cdot 1}$ .

$x - \frac{3}{4} = \pm \sqrt{i^{2} ({(\frac{1}{4})}^{2} \cdot 39)}$

Step 2.1.5

Rewrite $i^{2} {(\frac{1}{4})}^{2}$ as ${(\frac{i}{4})}^{2}$ .

$x - \frac{3}{4} = \pm \sqrt{{(\frac{i}{4})}^{2} \cdot 39}$

Step 2.2

Pull terms out from under the radical.

$x - \frac{3}{4} = \pm \frac{i}{4} \sqrt{39}$

Step 2.3

Combine $\frac{i}{4}$ and $\sqrt{39}$ .

$x - \frac{3}{4} = \pm \frac{i \sqrt{39}}{4}$

Step 3

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

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Step 3.1

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

$x - \frac{3}{4} = \frac{i \sqrt{39}}{4}$

Step 3.2

Add $\frac{3}{4}$ to both sides of the equation.

$x = \frac{i \sqrt{39}}{4} + \frac{3}{4}$

Step 3.3

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

$x - \frac{3}{4} = - \frac{i \sqrt{39}}{4}$

Step 3.4

Add $\frac{3}{4}$ to both sides of the equation.

$x = - \frac{i \sqrt{39}}{4} + \frac{3}{4}$

Step 3.5

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

$x = \frac{i \sqrt{39}}{4} + \frac{3}{4}, - \frac{i \sqrt{39}}{4} + \frac{3}{4}$