Precalculus Examples

$81 y^{4} + 1 = 18 y^{2}$

Step 1

Subtract $18 y^{2}$ from both sides of the equation.

$81 y^{4} + 1 - 18 y^{2} = 0$

Step 2

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

$81 u^{2} - 18 u + 1 = 0$

$u = y^{2}$

Step 3

Factor using the perfect square rule.

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Step 3.1

Rewrite $81 u^{2}$ as ${(9 u)}^{2}$ .

${(9 u)}^{2} - 18 u + 1 = 0$

Step 3.2

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

${(9 u)}^{2} - 18 u + 1^{2} = 0$

Step 3.3

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$18 u = 2 \cdot (9 u) \cdot 1$

Step 3.4

Rewrite the polynomial.

${(9 u)}^{2} - 2 \cdot (9 u) \cdot 1 + 1^{2} = 0$

Step 3.5

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = 9 u$ and $b = 1$ .

${(9 u - 1)}^{2} = 0$

Step 4

Set the $9 u - 1$ equal to $0$ .

$9 u - 1 = 0$

Step 5

Solve for $u$ .

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Step 5.1

Add $1$ to both sides of the equation.

$9 u = 1$

Step 5.2

Divide each term in $9 u = 1$ by $9$ and simplify.

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Step 5.2.1

Divide each term in $9 u = 1$ by $9$ .

$\frac{9 u}{9} = \frac{1}{9}$

Step 5.2.2

Simplify the left side.

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Step 5.2.2.1

Cancel the common factor of $9$ .

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Step 5.2.2.1.1

Cancel the common factor.

$\frac{9 u}{9} = \frac{1}{9}$

Step 5.2.2.1.2

Divide $u$ by $1$ .

$u = \frac{1}{9}$

Step 6

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

$y^{2} = \frac{1}{9}$

Step 7

Solve the equation for $y$ .

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Step 7.1

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

$y = \pm \sqrt{\frac{1}{9}}$

Step 7.2

Simplify $\pm \sqrt{\frac{1}{9}}$ .

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Step 7.2.1

Rewrite $\sqrt{\frac{1}{9}}$ as $\frac{\sqrt{1}}{\sqrt{9}}$ .

$y = \pm \frac{\sqrt{1}}{\sqrt{9}}$

Step 7.2.2

Any root of $1$ is $1$ .

$y = \pm \frac{1}{\sqrt{9}}$

Step 7.2.3

Simplify the denominator.

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Step 7.2.3.1

Rewrite $9$ as $3^{2}$ .

$y = \pm \frac{1}{\sqrt{3^{2}}}$

Step 7.2.3.2

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

$y = \pm \frac{1}{3}$

Step 7.3

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

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Step 7.3.1

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

$y = \frac{1}{3}$

Step 7.3.2

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

$y = - \frac{1}{3}$

Step 7.3.3

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

$y = \frac{1}{3}, - \frac{1}{3}$

Step 8

The result can be shown in multiple forms.

Exact Form:

$y = \frac{1}{3}, - \frac{1}{3}$

Decimal Form:

$y = 0.\overline{3}, - 0.\overline{3}$