Algebra Examples

4 x^{2} + 3 x - 1 = 0

$4 x^{2} + 3 x - 1 = 0$

Step 1

Use the quadratic formula to find the solutions.

\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Step 2

Substitute the values

a = 4

$a = 4$ ,

b = 3

$b = 3$ , and

c = - 1

$c = - 1$ into the quadratic formula and solve for

x

$x$ .

\frac{- 3 \pm \sqrt{3^{2} - 4 \cdot (4 \cdot - 1)}}{2 \cdot 4}

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

Step 3

Simplify.

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Step 3.1

Simplify the numerator.

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Step 3.1.1

Raise

3

$3$ to the power of

2

$2$ .

x = \frac{- 3 \pm \sqrt{9 - 4 \cdot 4 \cdot - 1}}{2 \cdot 4}

$x = \frac{- 3 \pm \sqrt{9 - 4 \cdot 4 \cdot - 1}}{2 \cdot 4}$

Step 3.1.2

Multiply

- 4 \cdot 4 \cdot - 1

$- 4 \cdot 4 \cdot - 1$ .

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Step 3.1.2.1

Multiply

- 4

$- 4$ by

4

$4$ .

x = \frac{- 3 \pm \sqrt{9 - 16 \cdot - 1}}{2 \cdot 4}

$x = \frac{- 3 \pm \sqrt{9 - 16 \cdot - 1}}{2 \cdot 4}$

Step 3.1.2.2

Multiply

- 16

$- 16$ by

- 1

$- 1$ .

x = \frac{- 3 \pm \sqrt{9 + 16}}{2 \cdot 4}

$x = \frac{- 3 \pm \sqrt{9 + 16}}{2 \cdot 4}$

x = \frac{- 3 \pm \sqrt{9 + 16}}{2 \cdot 4}

$x = \frac{- 3 \pm \sqrt{9 + 16}}{2 \cdot 4}$

Step 3.1.3

Add

9

$9$ and

16

$16$ .

x = \frac{- 3 \pm \sqrt{25}}{2 \cdot 4}

$x = \frac{- 3 \pm \sqrt{25}}{2 \cdot 4}$

Step 3.1.4

Rewrite

25

$25$ as

5^{2}

$5^{2}$ .

x = \frac{- 3 \pm \sqrt{5^{2}}}{2 \cdot 4}

$x = \frac{- 3 \pm \sqrt{5^{2}}}{2 \cdot 4}$

Step 3.1.5

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

x = \frac{- 3 \pm 5}{2 \cdot 4}

$x = \frac{- 3 \pm 5}{2 \cdot 4}$

x = \frac{- 3 \pm 5}{2 \cdot 4}

$x = \frac{- 3 \pm 5}{2 \cdot 4}$

Step 3.2

Multiply

2

$2$ by

4

$4$ .

x = \frac{- 3 \pm 5}{8}

$x = \frac{- 3 \pm 5}{8}$

x = \frac{- 3 \pm 5}{8}

$x = \frac{- 3 \pm 5}{8}$

Step 4

The final answer is the combination of both solutions.

x = \frac{1}{4}, - 1

$x = \frac{1}{4}, - 1$