Finite Math Examples

$\sqrt[4]{12 x^{2} - 35} = x$

Step 1

To remove the radical on the left side of the equation, raise both sides of the equation to the power of $4$ .

${\sqrt[4]{12 x^{2} - 35}}^{4} = x^{4}$

Step 2

Simplify each side of the equation.

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Step 2.1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt[4]{12 x^{2} - 35}$ as ${(12 x^{2} - 35)}^{\frac{1}{4}}$ .

${({(12 x^{2} - 35)}^{\frac{1}{4}})}^{4} = x^{4}$

Step 2.2

Simplify the left side.

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Step 2.2.1

Simplify ${({(12 x^{2} - 35)}^{\frac{1}{4}})}^{4}$ .

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Step 2.2.1.1

Multiply the exponents in ${({(12 x^{2} - 35)}^{\frac{1}{4}})}^{4}$ .

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Step 2.2.1.1.1

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

${(12 x^{2} - 35)}^{\frac{1}{4} \cdot 4} = x^{4}$

Step 2.2.1.1.2

Cancel the common factor of $4$ .

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Step 2.2.1.1.2.1

Cancel the common factor.

${(12 x^{2} - 35)}^{\frac{1}{4} \cdot 4} = x^{4}$

Step 2.2.1.1.2.2

Rewrite the expression.

${(12 x^{2} - 35)}^{1} = x^{4}$

Step 2.2.1.2

Simplify.

$12 x^{2} - 35 = x^{4}$

Step 3

Solve for $x$ .

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Step 3.1

Subtract $x^{4}$ from both sides of the equation.

$12 x^{2} - 35 - x^{4} = 0$

Step 3.2

Substitute $u = x^{2}$ into the equation. This will make the quadratic formula easy to use.

$- u^{2} + 12 u - 35 = 0$

$u = x^{2}$

Step 3.3

Factor the left side of the equation.

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Step 3.3.1

Factor $- 1$ out of $- u^{2} + 12 u - 35$ .

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Step 3.3.1.1

Factor $- 1$ out of $- u^{2}$ .

$- (u^{2}) + 12 u - 35 = 0$

Step 3.3.1.2

Factor $- 1$ out of $12 u$ .

$- (u^{2}) - (- 12 u) - 35 = 0$

Step 3.3.1.3

Rewrite $- 35$ as $- 1 (35)$ .

$- (u^{2}) - (- 12 u) - 1 \cdot 35 = 0$

Step 3.3.1.4

Factor $- 1$ out of $- (u^{2}) - (- 12 u)$ .

$- (u^{2} - 12 u) - 1 \cdot 35 = 0$

Step 3.3.1.5

Factor $- 1$ out of $- (u^{2} - 12 u) - 1 (35)$ .

$- (u^{2} - 12 u + 35) = 0$

Step 3.3.2

Factor.

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Step 3.3.2.1

Factor $u^{2} - 12 u + 35$ using the AC method.

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Step 3.3.2.1.1

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $35$ and whose sum is $- 12$ .

$- 7, - 5$

Step 3.3.2.1.2

Write the factored form using these integers.

$- ((u - 7) (u - 5)) = 0$

Step 3.3.2.2

Remove unnecessary parentheses.

$- (u - 7) (u - 5) = 0$

Step 3.4

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$u - 7 = 0$

$u - 5 = 0$

Step 3.5

Set $u - 7$ equal to $0$ and solve for $u$ .

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Step 3.5.1

Set $u - 7$ equal to $0$ .

$u - 7 = 0$

Step 3.5.2

Add $7$ to both sides of the equation.

$u = 7$

Step 3.6

Set $u - 5$ equal to $0$ and solve for $u$ .

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Step 3.6.1

Set $u - 5$ equal to $0$ .

$u - 5 = 0$

Step 3.6.2

Add $5$ to both sides of the equation.

$u = 5$

Step 3.7

The final solution is all the values that make $- (u - 7) (u - 5) = 0$ true.

$u = 7, 5$

Step 3.8

Substitute the real value of $u = x^{2}$ back into the solved equation.

$x^{2} = 7$

${(x^{2})}^{1} = 5$

Step 3.9

Solve the first equation for $x$ .

$x^{2} = 7$

Step 3.10

Solve the equation for $x$ .

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Step 3.10.1

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

$x = \pm \sqrt{7}$

Step 3.10.2

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

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Step 3.10.2.1

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

$x = \sqrt{7}$

Step 3.10.2.2

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

$x = - \sqrt{7}$

Step 3.10.2.3

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

$x = \sqrt{7}, - \sqrt{7}$

Step 3.11

Solve the second equation for $x$ .

${(x^{2})}^{1} = 5$

Step 3.12

Solve the equation for $x$ .

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Step 3.12.1

Remove parentheses.

$x^{2} = 5$

Step 3.12.2

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

$x = \pm \sqrt{5}$

Step 3.12.3

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

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Step 3.12.3.1

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

$x = \sqrt{5}$

Step 3.12.3.2

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

$x = - \sqrt{5}$

Step 3.12.3.3

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

$x = \sqrt{5}, - \sqrt{5}$

Step 3.13

The solution to $- x^{4} + 12 x^{2} - 35 = 0$ is $x = \sqrt{7}, - \sqrt{7}, \sqrt{5}, - \sqrt{5}$ .

$x = \sqrt{7}, - \sqrt{7}, \sqrt{5}, - \sqrt{5}$

Step 4

Exclude the solutions that do not make $\sqrt[4]{12 x^{2} - 35} = x$ true.

$x = \sqrt{7}, \sqrt{5}$

Step 5

The result can be shown in multiple forms.

Exact Form:

$x = \sqrt{7}, \sqrt{5}$

Decimal Form:

$x = 2.64575131 \dots, 2.23606797 \dots$