Precalculus Examples

$26 x^{2} + 1 = 10 x$

Step 1

Subtract $10 x$ from both sides of the equation.

$26 x^{2} + 1 - 10 x = 0$

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 $a = 26$ , $b = - 10$ , and $c = 1$ into the quadratic formula and solve for $x$ .

$\frac{10 \pm \sqrt{{(- 10)}^{2} - 4 \cdot (26 \cdot 1)}}{2 \cdot 26}$

Step 4

Simplify.

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Step 4.1

Simplify the numerator.

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Step 4.1.1

Raise $- 10$ to the power of $2$ .

$x = \frac{10 \pm \sqrt{100 - 4 \cdot 26 \cdot 1}}{2 \cdot 26}$

Step 4.1.2

Multiply $- 4 \cdot 26 \cdot 1$ .

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Step 4.1.2.1

Multiply $- 4$ by $26$ .

$x = \frac{10 \pm \sqrt{100 - 104 \cdot 1}}{2 \cdot 26}$

Step 4.1.2.2

Multiply $- 104$ by $1$ .

$x = \frac{10 \pm \sqrt{100 - 104}}{2 \cdot 26}$

Step 4.1.3

Subtract $104$ from $100$ .

$x = \frac{10 \pm \sqrt{- 4}}{2 \cdot 26}$

Step 4.1.4

Rewrite $- 4$ as $- 1 (4)$ .

$x = \frac{10 \pm \sqrt{- 1 \cdot 4}}{2 \cdot 26}$

Step 4.1.5

Rewrite $\sqrt{- 1 (4)}$ as $\sqrt{- 1} \cdot \sqrt{4}$ .

$x = \frac{10 \pm \sqrt{- 1} \cdot \sqrt{4}}{2 \cdot 26}$

Step 4.1.6

Rewrite $\sqrt{- 1}$ as $i$ .

$x = \frac{10 \pm i \cdot \sqrt{4}}{2 \cdot 26}$

Step 4.1.7

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

$x = \frac{10 \pm i \cdot \sqrt{2^{2}}}{2 \cdot 26}$

Step 4.1.8

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

$x = \frac{10 \pm i \cdot 2}{2 \cdot 26}$

Step 4.1.9

Move $2$ to the left of $i$ .

$x = \frac{10 \pm 2 i}{2 \cdot 26}$

Step 4.2

Multiply $2$ by $26$ .

$x = \frac{10 \pm 2 i}{52}$

Step 4.3

Simplify $\frac{10 \pm 2 i}{52}$ .

$x = \frac{5 \pm i}{26}$