Solve for n t = square root of (n+3)/n

$t = \sqrt{\frac{n + 3}{n}}$

Step 1

Rewrite the equation as $\sqrt{\frac{n + 3}{n}} = t$ .

$\sqrt{\frac{n + 3}{n}} = t$

Step 2

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

${\sqrt{\frac{n + 3}{n}}}^{2} = t^{2}$

Step 3

Simplify each side of the equation.

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

${({(\frac{n + 3}{n})}^{\frac{1}{2}})}^{2} = t^{2}$

Simplify the left side.

Simplify ${({(\frac{n + 3}{n})}^{\frac{1}{2}})}^{2}$ .

Multiply the exponents in ${({(\frac{n + 3}{n})}^{\frac{1}{2}})}^{2}$ .

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

${(\frac{n + 3}{n})}^{\frac{1}{2} \cdot 2} = t^{2}$

Cancel the common factor of $2$ .

Cancel the common factor.

${(\frac{n + 3}{n})}^{\frac{1}{2} \cdot 2} = t^{2}$

Rewrite the expression.

${(\frac{n + 3}{n})}^{1} = t^{2}$

Simplify.

$\frac{n + 3}{n} = t^{2}$

Step 4

Solve for $n$ .

Find the LCD of the terms in the equation.

Finding the LCD of a list of values is the same as finding the LCM of the denominators of those values.

$n, 1$

The LCM of one and any expression is the expression.

$n$

Multiply each term in $\frac{n + 3}{n} = t^{2}$ by $n$ to eliminate the fractions.

Multiply each term in $\frac{n + 3}{n} = t^{2}$ by $n$ .

$\frac{n + 3}{n} n = t^{2} n$

Simplify the left side.

Cancel the common factor of $n$ .

Cancel the common factor.

$\frac{n + 3}{n} n = t^{2} n$

Rewrite the expression.

$n + 3 = t^{2} n$

Solve the equation.

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

$n + 3 - t^{2} n = 0$

Subtract $3$ from both sides of the equation.

$n - t^{2} n = - 3$

Factor $n$ out of $n - t^{2} n$ .

Factor $n$ out of $n^{1}$ .

$n \cdot 1 - t^{2} n = - 3$

Factor $n$ out of $- t^{2} n$ .

$n \cdot 1 + n (- t^{2}) = - 3$

Factor $n$ out of $n \cdot 1 + n (- t^{2})$ .

$n (1 - t^{2}) = - 3$

Rewrite $1$ as $1^{2}$ .

$n (1^{2} - t^{2}) = - 3$

Factor.

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 1$ and $b = t$ .

$n ((1 + t) (1 - t)) = - 3$

Remove unnecessary parentheses.

$n (1 + t) (1 - t) = - 3$

Divide each term in $n (1 + t) (1 - t) = - 3$ by $(1 + t) (1 - t)$ and simplify.

Divide each term in $n (1 + t) (1 - t) = - 3$ by $(1 + t) (1 - t)$ .

$\frac{n (1 + t) (1 - t)}{(1 + t) (1 - t)} = \frac{- 3}{(1 + t) (1 - t)}$

Simplify the left side.

Cancel the common factor of $1 + t$ .

Cancel the common factor.

$\frac{n (1 + t) (1 - t)}{(1 + t) (1 - t)} = \frac{- 3}{(1 + t) (1 - t)}$

Rewrite the expression.

$\frac{n (1 - t)}{1 - t} = \frac{- 3}{(1 + t) (1 - t)}$

Cancel the common factor of $1 - t$ .

Cancel the common factor.

$\frac{n (1 - t)}{1 - t} = \frac{- 3}{(1 + t) (1 - t)}$

Divide $n$ by $1$ .

$n = \frac{- 3}{(1 + t) (1 - t)}$

Simplify the right side.

Move the negative in front of the fraction.

$n = - \frac{3}{(1 + t) (1 - t)}$

Solve for n t = square root of (n+3)/n

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