Basic Math Examples

$(16 - a^{2}) \cdot \sqrt{a + 3} = 0$

Step 1

Apply the distributive property.

$16 \sqrt{a + 3} - a^{2} \sqrt{a + 3} = 0$

Step 2

Simplify the left side.

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Step 2.1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{a + 3}$ as ${(a + 3)}^{\frac{1}{2}}$ .

$16 {(a + 3)}^{\frac{1}{2}} - a^{2} \sqrt{a + 3} = 0$

Step 2.2

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{a + 3}$ as ${(a + 3)}^{\frac{1}{2}}$ .

$16 {(a + 3)}^{\frac{1}{2}} - a^{2} {(a + 3)}^{\frac{1}{2}} = 0$

Step 3

Factor the left side of the equation.

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Step 3.1

Add parentheses.

$16 {(a + 3)}^{\frac{1}{2}} - (a^{2} {(a + 3)}^{\frac{1}{2}}) = 0$

Step 3.2

Let $u = {(a + 3)}^{\frac{1}{2}}$ . Substitute $u$ for all occurrences of ${(a + 3)}^{\frac{1}{2}}$ .

$16 u - a^{2} u = 0$

Step 3.3

Factor $u$ out of $16 u - a^{2} u$ .

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Step 3.3.1

Factor $u$ out of $16 u$ .

$u \cdot 16 - a^{2} u = 0$

Step 3.3.2

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

$u \cdot 16 + u (- a^{2}) = 0$

Step 3.3.3

Factor $u$ out of $u \cdot 16 + u (- a^{2})$ .

$u (16 - a^{2}) = 0$

Step 3.4

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

$u (4^{2} - a^{2}) = 0$

Step 3.5

Factor.

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Step 3.5.1

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 4$ and $b = a$ .

$u ((4 + a) (4 - a)) = 0$

Step 3.5.2

Remove unnecessary parentheses.

$u (4 + a) (4 - a) = 0$

Step 3.6

Replace all occurrences of $u$ with ${(a + 3)}^{\frac{1}{2}}$ .

${(a + 3)}^{\frac{1}{2}} (4 + a) (4 - a) = 0$

Step 4

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

${(a + 3)}^{\frac{1}{2}} = 0$

$4 + a = 0$

$4 - a = 0$

Step 5

Set ${(a + 3)}^{\frac{1}{2}}$ equal to $0$ and solve for $a$ .

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Step 5.1

Set ${(a + 3)}^{\frac{1}{2}}$ equal to $0$ .

${(a + 3)}^{\frac{1}{2}} = 0$

Step 5.2

Solve ${(a + 3)}^{\frac{1}{2}} = 0$ for $a$ .

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Step 5.2.1

Set the $a + 3$ equal to $0$ .

$a + 3 = 0$

Step 5.2.2

Subtract $3$ from both sides of the equation.

$a = - 3$

Step 6

Set $4 + a$ equal to $0$ and solve for $a$ .

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Step 6.1

Set $4 + a$ equal to $0$ .

$4 + a = 0$

Step 6.2

Subtract $4$ from both sides of the equation.

$a = - 4$

Step 7

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

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Step 7.1

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

$4 - a = 0$

Step 7.2

Solve $4 - a = 0$ for $a$ .

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Step 7.2.1

Subtract $4$ from both sides of the equation.

$- a = - 4$

Step 7.2.2

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

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Step 7.2.2.1

Divide each term in $- a = - 4$ by $- 1$ .

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

Step 7.2.2.2

Simplify the left side.

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Step 7.2.2.2.1

Dividing two negative values results in a positive value.

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

Step 7.2.2.2.2

Divide $a$ by $1$ .

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

Step 7.2.2.3

Simplify the right side.

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Step 7.2.2.3.1

Divide $- 4$ by $- 1$ .

$a = 4$

Step 8

The final solution is all the values that make ${(a + 3)}^{\frac{1}{2}} (4 + a) (4 - a) = 0$ true.

$a = - 3, - 4, 4$

Step 9

Exclude the solutions that do not make $(16 - a^{2}) \cdot \sqrt{a + 3} = 0$ true.

$a = - 3, 4$