Basic Math Examples

$y^{4} - 20 y^{2} + 64 = 0$

Step 1

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

$u^{2} - 20 u + 64 = 0$

$u = y^{2}$

Step 2

Factor $u^{2} - 20 u + 64$ using the AC method.

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Step 2.1

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 $64$ and whose sum is $- 20$ .

$- 16, - 4$

Step 2.2

Write the factored form using these integers.

$(u - 16) (u - 4) = 0$

Step 3

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

$u - 16 = 0$

$u - 4 = 0$

Step 4

Set $u - 16$ equal to $0$ and solve for $u$ .

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Step 4.1

Set $u - 16$ equal to $0$ .

$u - 16 = 0$

Step 4.2

Add $16$ to both sides of the equation.

$u = 16$

Step 5

Set $u - 4$ equal to $0$ and solve for $u$ .

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Step 5.1

Set $u - 4$ equal to $0$ .

$u - 4 = 0$

Step 5.2

Add $4$ to both sides of the equation.

$u = 4$

Step 6

The final solution is all the values that make $(u - 16) (u - 4) = 0$ true.

$u = 16, 4$

Step 7

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

$y^{2} = 16$

${(y^{2})}^{1} = 4$

Step 8

Solve the first equation for $y$ .

$y^{2} = 16$

Step 9

Solve the equation for $y$ .

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Step 9.1

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

$y = \pm \sqrt{16}$

Step 9.2

Simplify $\pm \sqrt{16}$ .

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Step 9.2.1

Rewrite $16$ as $4^{2}$ .

$y = \pm \sqrt{4^{2}}$

Step 9.2.2

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

$y = \pm 4$

Step 9.3

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

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Step 9.3.1

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

$y = 4$

Step 9.3.2

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

$y = - 4$

Step 9.3.3

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

$y = 4, - 4$

Step 10

Solve the second equation for $y$ .

${(y^{2})}^{1} = 4$

Step 11

Solve the equation for $y$ .

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Step 11.1

Remove parentheses.

$y^{2} = 4$

Step 11.2

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

$y = \pm \sqrt{4}$

Step 11.3

Simplify $\pm \sqrt{4}$ .

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Step 11.3.1

Rewrite $4$ as $2^{2}$ .

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

Step 11.3.2

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

$y = \pm 2$

Step 11.4

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

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Step 11.4.1

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

$y = 2$

Step 11.4.2

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

$y = - 2$

Step 11.4.3

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

$y = 2, - 2$

Step 12

The solution to $y^{4} - 20 y^{2} + 64 = 0$ is $y = 4, - 4, 2, - 2$ .

$y = 4, - 4, 2, - 2$