Algebra Examples

$y^{4} = 2 - y^{2}$

Step 1

Add $y^{2}$ to both sides of the equation.

$y^{4} + y^{2} = 2$

Step 2

Subtract $2$ from both sides of the equation.

$y^{4} + y^{2} - 2 = 0$

Step 3

Factor the left side of the equation.

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Step 3.1

Rewrite $y^{4}$ as ${(y^{2})}^{2}$ .

${(y^{2})}^{2} + y^{2} - 2 = 0$

Step 3.2

Let $u = y^{2}$ . Substitute $u$ for all occurrences of $y^{2}$ .

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

Step 3.3

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

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Step 3.3.1

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 $- 2$ and whose sum is $1$ .

$- 1, 2$

Step 3.3.2

Write the factored form using these integers.

$(u - 1) (u + 2) = 0$

Step 3.4

Replace all occurrences of $u$ with $y^{2}$ .

$(y^{2} - 1) (y^{2} + 2) = 0$

Step 3.5

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

$(y^{2} - 1^{2}) (y^{2} + 2) = 0$

Step 3.6

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = y$ and $b = 1$ .

$(y + 1) (y - 1) (y^{2} + 2) = 0$

Step 4

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

$y + 1 = 0$

$y - 1 = 0$

$y^{2} + 2 = 0$

Step 5

Set $y + 1$ equal to $0$ and solve for $y$ .

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Step 5.1

Set $y + 1$ equal to $0$ .

$y + 1 = 0$

Step 5.2

Subtract $1$ from both sides of the equation.

$y = - 1$

Step 6

Set $y - 1$ equal to $0$ and solve for $y$ .

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Step 6.1

Set $y - 1$ equal to $0$ .

$y - 1 = 0$

Step 6.2

Add $1$ to both sides of the equation.

$y = 1$

Step 7

Set $y^{2} + 2$ equal to $0$ and solve for $y$ .

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Step 7.1

Set $y^{2} + 2$ equal to $0$ .

$y^{2} + 2 = 0$

Step 7.2

Solve $y^{2} + 2 = 0$ for $y$ .

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Step 7.2.1

Subtract $2$ from both sides of the equation.

$y^{2} = - 2$

Step 7.2.2

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

$y = \pm \sqrt{- 2}$

Step 7.2.3

Simplify $\pm \sqrt{- 2}$ .

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Step 7.2.3.1

Rewrite $- 2$ as $- 1 (2)$ .

$y = \pm \sqrt{- 1 (2)}$

Step 7.2.3.2

Rewrite $\sqrt{- 1 (2)}$ as $\sqrt{- 1} \cdot \sqrt{2}$ .

$y = \pm \sqrt{- 1} \cdot \sqrt{2}$

Step 7.2.3.3

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

$y = \pm i \sqrt{2}$

Step 7.2.4

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

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Step 7.2.4.1

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

$y = i \sqrt{2}$

Step 7.2.4.2

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

$y = - i \sqrt{2}$

Step 7.2.4.3

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

$y = i \sqrt{2}, - i \sqrt{2}$

Step 8

The final solution is all the values that make $(y + 1) (y - 1) (y^{2} + 2) = 0$ true.

$y = - 1, 1, i \sqrt{2}, - i \sqrt{2}$