How do you solve $\sqrt{2 n - 88} = \sqrt{\frac{n}{6}}$ $sqrt(2n-88)=sqrt(n/6)$ ?

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How do you solve $\sqrt{2 n - 88} = \sqrt{\frac{n}{6}}$ $sqrt(2n-88)=sqrt(n/6)$ ?

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Answer 1 · 2016-10-31T10:03:22

The solution of the equation is $n = 48$ $n = 48$ .

Explanation:

First, you must get rid of the radicals. In this case, that is achieved by squaring both sides of the equation.

${(\sqrt{2 n - 88})}^{2} = {(\sqrt{\frac{n}{6}})}^{2}$ $(sqrt(2n - 88))^2 = (sqrt(n/6))^2$

$2 n - 88 = \frac{n}{6}$ $2n - 88 = n/6$

Now use inverse operations to solve for $n$ $n$ .

$(2 n - 88) 6 = (\frac{n}{6}) 6$ $(2n - 88)6 = (n/6)6$

$12 n - 528 = n$ $12n - 528 = n$
$12 n - 12 n - 528 = n - 12 n$ $12n - 12n - 528 = n - 12n$
$- 528 = - 11 n$ $-528 = -11n$
$\frac{- 528}{- 11} = \frac{- 11 n}{- 11}$ $(-528)/-11 = (-11n)/-11$

$48 = n$ $48 = n$

Now we must check the solution to be sure that it is not extraneous. This step absolutely cannot be skipped when solving radical equations! Remember to check a solution, we put it in place of the variable in the original equation and simplify each side of the equation to see if it makes a true statement.

$\sqrt{2 \cdot 48 - 88} = \sqrt{\frac{48}{6}}$ $sqrt(2*48 - 88) = sqrt(48/6)$

$\sqrt{96 - 88} = \sqrt{8}$ $sqrt(96 - 88) = sqrt(8)$
$\sqrt{8} = 2 \sqrt{2}$ $sqrt(8) = 2sqrt(2)$
$2 \sqrt{2} = 2 \sqrt{2}$ $2sqrt(2) = 2sqrt(2)$

Since the two sides of the equation simplify to be equal, $48$ $48$ is the solution of the equation.

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How do you solve $\sqrt{2 n - 88} = \sqrt{\frac{n}{6}}$ $sqrt(2n-88)=sqrt(n/6)$ ?

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How do you solve $\sqrt{2 n - 88} = \sqrt{\frac{n}{6}}$ $sqrt(2n-88)=sqrt(n/6)$ ?