Finite Math Examples

$x^{2} - 2 \sqrt{5} x + 5 = 0$

Step 1

Use the quadratic formula to find the solutions.

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

Step 2

Substitute the values $a = 1$ , $b = - 2 \sqrt{5}$ , and $c = 5$ into the quadratic formula and solve for $x$ .

$\frac{2 \sqrt{5} \pm \sqrt{{(- 2 \sqrt{5})}^{2} - 4 \cdot (1 \cdot 5)}}{2 \cdot 1}$

Step 3

Simplify.

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Step 3.1

Simplify the numerator.

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Step 3.1.1

Apply the product rule to $- 2 \sqrt{5}$ .

$x = \frac{2 \sqrt{5} \pm \sqrt{{(- 2)}^{2} {\sqrt{5}}^{2} - 4 \cdot 1 \cdot 5}}{2 \cdot 1}$

Step 3.1.2

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

$x = \frac{2 \sqrt{5} \pm \sqrt{4 {\sqrt{5}}^{2} - 4 \cdot 1 \cdot 5}}{2 \cdot 1}$

Step 3.1.3

Rewrite ${\sqrt{5}}^{2}$ as $5$ .

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Step 3.1.3.1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{5}$ as $5^{\frac{1}{2}}$ .

$x = \frac{2 \sqrt{5} \pm \sqrt{4 {(5^{\frac{1}{2}})}^{2} - 4 \cdot 1 \cdot 5}}{2 \cdot 1}$

Step 3.1.3.2

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$x = \frac{2 \sqrt{5} \pm \sqrt{4 \cdot 5^{\frac{1}{2} \cdot 2} - 4 \cdot 1 \cdot 5}}{2 \cdot 1}$

Step 3.1.3.3

Combine $\frac{1}{2}$ and $2$ .

$x = \frac{2 \sqrt{5} \pm \sqrt{4 \cdot 5^{\frac{2}{2}} - 4 \cdot 1 \cdot 5}}{2 \cdot 1}$

Step 3.1.3.4

Cancel the common factor of $2$ .

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Step 3.1.3.4.1

Cancel the common factor.

$x = \frac{2 \sqrt{5} \pm \sqrt{4 \cdot 5^{\frac{2}{2}} - 4 \cdot 1 \cdot 5}}{2 \cdot 1}$

Step 3.1.3.4.2

Rewrite the expression.

$x = \frac{2 \sqrt{5} \pm \sqrt{4 \cdot 5 - 4 \cdot 1 \cdot 5}}{2 \cdot 1}$

Step 3.1.3.5

Evaluate the exponent.

$x = \frac{2 \sqrt{5} \pm \sqrt{4 \cdot 5 - 4 \cdot 1 \cdot 5}}{2 \cdot 1}$

Step 3.1.4

Multiply $4$ by $5$ .

$x = \frac{2 \sqrt{5} \pm \sqrt{20 - 4 \cdot 1 \cdot 5}}{2 \cdot 1}$

Step 3.1.5

Multiply $- 4 \cdot 1 \cdot 5$ .

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Step 3.1.5.1

Multiply $- 4$ by $1$ .

$x = \frac{2 \sqrt{5} \pm \sqrt{20 - 4 \cdot 5}}{2 \cdot 1}$

Step 3.1.5.2

Multiply $- 4$ by $5$ .

$x = \frac{2 \sqrt{5} \pm \sqrt{20 - 20}}{2 \cdot 1}$

Step 3.1.6

Subtract $20$ from $20$ .

$x = \frac{2 \sqrt{5} \pm \sqrt{0}}{2 \cdot 1}$

Step 3.1.7

Rewrite $0$ as $0^{2}$ .

$x = \frac{2 \sqrt{5} \pm \sqrt{0^{2}}}{2 \cdot 1}$

Step 3.1.8

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

$x = \frac{2 \sqrt{5} \pm 0}{2 \cdot 1}$

Step 3.1.9

$2 \sqrt{5}$ plus or minus $0$ is $2 \sqrt{5}$ .

$x = \frac{2 \sqrt{5}}{2 \cdot 1}$

Step 3.2

Multiply $2$ by $1$ .

$x = \frac{2 \sqrt{5}}{2}$

Step 3.3

Cancel the common factor of $2$ .

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Step 3.3.1

Cancel the common factor.

$x = \frac{2 \sqrt{5}}{2}$

Step 3.3.2

Divide $\sqrt{5}$ by $1$ .

$x = \sqrt{5}$

Step 4

The final answer is the combination of both solutions.

$x = \sqrt{5}$ Double roots