Basic Math Examples

$3 \sqrt{10 - k} - k = 0$

Step 1

Add $k$ to both sides of the equation.

$3 \sqrt{10 - k} = k$

Step 2

To remove the radical on the left side of the equation, square both sides of the equation.

${(3 \sqrt{10 - k})}^{2} = k^{2}$

Step 3

Simplify each side of the equation.

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Step 3.1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{10 - k}$ as ${(10 - k)}^{\frac{1}{2}}$ .

${(3 {(10 - k)}^{\frac{1}{2}})}^{2} = k^{2}$

Step 3.2

Simplify the left side.

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Step 3.2.1

Simplify ${(3 {(10 - k)}^{\frac{1}{2}})}^{2}$ .

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Step 3.2.1.1

Apply the product rule to $3 {(10 - k)}^{\frac{1}{2}}$ .

$3^{2} {({(10 - k)}^{\frac{1}{2}})}^{2} = k^{2}$

Step 3.2.1.2

Raise $3$ to the power of $2$ .

$9 {({(10 - k)}^{\frac{1}{2}})}^{2} = k^{2}$

Step 3.2.1.3

Multiply the exponents in ${({(10 - k)}^{\frac{1}{2}})}^{2}$ .

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Step 3.2.1.3.1

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$9 {(10 - k)}^{\frac{1}{2} \cdot 2} = k^{2}$

Step 3.2.1.3.2

Cancel the common factor of $2$ .

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Step 3.2.1.3.2.1

Cancel the common factor.

$9 {(10 - k)}^{\frac{1}{2} \cdot 2} = k^{2}$

Step 3.2.1.3.2.2

Rewrite the expression.

$9 {(10 - k)}^{1} = k^{2}$

Step 3.2.1.4

Simplify.

$9 (10 - k) = k^{2}$

Step 3.2.1.5

Apply the distributive property.

$9 \cdot 10 + 9 (- k) = k^{2}$

Step 3.2.1.6

Multiply.

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Step 3.2.1.6.1

Multiply $9$ by $10$ .

$90 + 9 (- k) = k^{2}$

Step 3.2.1.6.2

Multiply $- 1$ by $9$ .

$90 - 9 k = k^{2}$

Step 4

Solve for $k$ .

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Step 4.1

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

$90 - 9 k - k^{2} = 0$

Step 4.2

Factor the left side of the equation.

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Step 4.2.1

Factor $- 1$ out of $90 - 9 k - k^{2}$ .

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Step 4.2.1.1

Reorder the expression.

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Step 4.2.1.1.1

Move $90$ .

$- 9 k - k^{2} + 90 = 0$

Step 4.2.1.1.2

Reorder $- 9 k$ and $- k^{2}$ .

$- k^{2} - 9 k + 90 = 0$

Step 4.2.1.2

Factor $- 1$ out of $- k^{2}$ .

$- (k^{2}) - 9 k + 90 = 0$

Step 4.2.1.3

Factor $- 1$ out of $- 9 k$ .

$- (k^{2}) - (9 k) + 90 = 0$

Step 4.2.1.4

Rewrite $90$ as $- 1 (- 90)$ .

$- (k^{2}) - (9 k) - 1 \cdot - 90 = 0$

Step 4.2.1.5

Factor $- 1$ out of $- (k^{2}) - (9 k)$ .

$- (k^{2} + 9 k) - 1 \cdot - 90 = 0$

Step 4.2.1.6

Factor $- 1$ out of $- (k^{2} + 9 k) - 1 (- 90)$ .

$- (k^{2} + 9 k - 90) = 0$

Step 4.2.2

Factor.

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Step 4.2.2.1

Factor $k^{2} + 9 k - 90$ using the AC method.

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Step 4.2.2.1.1

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 $- 90$ and whose sum is $9$ .

$- 6, 15$

Step 4.2.2.1.2

Write the factored form using these integers.

$- ((k - 6) (k + 15)) = 0$

Step 4.2.2.2

Remove unnecessary parentheses.

$- (k - 6) (k + 15) = 0$

Step 4.3

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

$k - 6 = 0$

$k + 15 = 0$

Step 4.4

Set $k - 6$ equal to $0$ and solve for $k$ .

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Step 4.4.1

Set $k - 6$ equal to $0$ .

$k - 6 = 0$

Step 4.4.2

Add $6$ to both sides of the equation.

$k = 6$

Step 4.5

Set $k + 15$ equal to $0$ and solve for $k$ .

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Step 4.5.1

Set $k + 15$ equal to $0$ .

$k + 15 = 0$

Step 4.5.2

Subtract $15$ from both sides of the equation.

$k = - 15$

Step 4.6

The final solution is all the values that make $- (k - 6) (k + 15) = 0$ true.

$k = 6, - 15$

Step 5

Exclude the solutions that do not make $3 \sqrt{10 - k} - k = 0$ true.

$k = 6$