Basic Math Examples

$4 a^{4} - 65 a^{2} + 16 = 0$

Step 1

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

$4 u^{2} - 65 u + 16 = 0$

$u = a^{2}$

Step 2

Factor by grouping.

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Step 2.1

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = 4 \cdot 16 = 64$ and whose sum is $b = - 65$ .

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Step 2.1.1

Factor $- 65$ out of $- 65 u$ .

$4 u^{2} - 65 u + 16 = 0$

Step 2.1.2

Rewrite $- 65$ as $- 1$ plus $- 64$

$4 u^{2} + (- 1 - 64) u + 16 = 0$

Step 2.1.3

Apply the distributive property.

$4 u^{2} - 1 u - 64 u + 16 = 0$

Step 2.2

Factor out the greatest common factor from each group.

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Step 2.2.1

Group the first two terms and the last two terms.

$(4 u^{2} - 1 u) - 64 u + 16 = 0$

Step 2.2.2

Factor out the greatest common factor (GCF) from each group.

$u (4 u - 1) - 16 (4 u - 1) = 0$

Step 2.3

Factor the polynomial by factoring out the greatest common factor, $4 u - 1$ .

$(4 u - 1) (u - 16) = 0$

Step 3

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

$4 u - 1 = 0$

$u - 16 = 0$

Step 4

Set $4 u - 1$ equal to $0$ and solve for $u$ .

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Step 4.1

Set $4 u - 1$ equal to $0$ .

$4 u - 1 = 0$

Step 4.2

Solve $4 u - 1 = 0$ for $u$ .

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Step 4.2.1

Add $1$ to both sides of the equation.

$4 u = 1$

Step 4.2.2

Divide each term in $4 u = 1$ by $4$ and simplify.

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Step 4.2.2.1

Divide each term in $4 u = 1$ by $4$ .

$\frac{4 u}{4} = \frac{1}{4}$

Step 4.2.2.2

Simplify the left side.

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Step 4.2.2.2.1

Cancel the common factor of $4$ .

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Step 4.2.2.2.1.1

Cancel the common factor.

$\frac{4 u}{4} = \frac{1}{4}$

Step 4.2.2.2.1.2

Divide $u$ by $1$ .

$u = \frac{1}{4}$

Step 5

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

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Step 5.1

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

$u - 16 = 0$

Step 5.2

Add $16$ to both sides of the equation.

$u = 16$

Step 6

The final solution is all the values that make $(4 u - 1) (u - 16) = 0$ true.

$u = \frac{1}{4}, 16$

Step 7

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

$a^{2} = \frac{1}{4}$

${(a^{2})}^{1} = 16$

Step 8

Solve the first equation for $a$ .

$a^{2} = \frac{1}{4}$

Step 9

Solve the equation for $a$ .

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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.

$a = \pm \sqrt{\frac{1}{4}}$

Step 9.2

Simplify $\pm \sqrt{\frac{1}{4}}$ .

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Step 9.2.1

Rewrite $\sqrt{\frac{1}{4}}$ as $\frac{\sqrt{1}}{\sqrt{4}}$ .

$a = \pm \frac{\sqrt{1}}{\sqrt{4}}$

Step 9.2.2

Any root of $1$ is $1$ .

$a = \pm \frac{1}{\sqrt{4}}$

Step 9.2.3

Simplify the denominator.

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Step 9.2.3.1

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

$a = \pm \frac{1}{\sqrt{2^{2}}}$

Step 9.2.3.2

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

$a = \pm \frac{1}{2}$

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.

$a = \frac{1}{2}$

Step 9.3.2

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

$a = - \frac{1}{2}$

Step 9.3.3

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

$a = \frac{1}{2}, - \frac{1}{2}$

Step 10

Solve the second equation for $a$ .

${(a^{2})}^{1} = 16$

Step 11

Solve the equation for $a$ .

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Step 11.1

Remove parentheses.

$a^{2} = 16$

Step 11.2

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

$a = \pm \sqrt{16}$

Step 11.3

Simplify $\pm \sqrt{16}$ .

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Step 11.3.1

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

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

Step 11.3.2

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

$a = \pm 4$

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.

$a = 4$

Step 11.4.2

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

$a = - 4$

Step 11.4.3

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

$a = 4, - 4$

Step 12

The solution to $4 a^{4} - 65 a^{2} + 16 = 0$ is $a = \frac{1}{2}, - \frac{1}{2}, 4, - 4$ .

$a = \frac{1}{2}, - \frac{1}{2}, 4, - 4$

Step 13

The result can be shown in multiple forms.

Exact Form:

$a = \frac{1}{2}, - \frac{1}{2}, 4, - 4$

Decimal Form:

$a = 0.5, - 0.5, 4, - 4$