Solve for g f(4g^2-9)=n

$f (4 g^{2} - 9) = n$

Step 1

Divide each term in $f (4 g^{2} - 9) = n$ by $f$ and simplify.

Divide each term in $f (4 g^{2} - 9) = n$ by $f$ .

$\frac{f (4 g^{2} - 9)}{f} = \frac{n}{f}$

Simplify the left side.

Cancel the common factor of $f$ .

Cancel the common factor.

$\frac{f (4 g^{2} - 9)}{f} = \frac{n}{f}$

Divide $4 g^{2} - 9$ by $1$ .

$4 g^{2} - 9 = \frac{n}{f}$

Step 2

Add $9$ to both sides of the equation.

$4 g^{2} = \frac{n}{f} + 9$

Step 3

Divide each term in $4 g^{2} = \frac{n}{f} + 9$ by $4$ and simplify.

Divide each term in $4 g^{2} = \frac{n}{f} + 9$ by $4$ .

$\frac{4 g^{2}}{4} = \frac{\frac{n}{f}}{4} + \frac{9}{4}$

Simplify the left side.

Cancel the common factor of $4$ .

Cancel the common factor.

$\frac{4 g^{2}}{4} = \frac{\frac{n}{f}}{4} + \frac{9}{4}$

Divide $g^{2}$ by $1$ .

$g^{2} = \frac{\frac{n}{f}}{4} + \frac{9}{4}$

Simplify the right side.

Simplify each term.

Multiply the numerator by the reciprocal of the denominator.

$g^{2} = \frac{n}{f} \cdot \frac{1}{4} + \frac{9}{4}$

Multiply $\frac{n}{f}$ by $\frac{1}{4}$ .

$g^{2} = \frac{n}{f \cdot 4} + \frac{9}{4}$

Move $4$ to the left of $f$ .

$g^{2} = \frac{n}{4 f} + \frac{9}{4}$

Step 4

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

$g = \pm \sqrt{\frac{n}{4 f} + \frac{9}{4}}$

Step 5

Simplify $\pm \sqrt{\frac{n}{4 f} + \frac{9}{4}}$ .

To write $\frac{9}{4}$ as a fraction with a common denominator, multiply by $\frac{f}{f}$ .

$g = \pm \sqrt{\frac{n}{4 f} + \frac{9}{4} \cdot \frac{f}{f}}$

Multiply $\frac{9}{4}$ by $\frac{f}{f}$ .

$g = \pm \sqrt{\frac{n}{4 f} + \frac{9 f}{4 f}}$

Combine the numerators over the common denominator.

$g = \pm \sqrt{\frac{n + 9 f}{4 f}}$

Rewrite $\frac{n + 9 f}{4 f}$ as ${(\frac{1}{2})}^{2} \frac{n + 9 f}{f}$ .

Factor the perfect power $1^{2}$ out of $n + 9 f$ .

$g = \pm \sqrt{\frac{1^{2} (n + 9 f)}{4 f}}$

Factor the perfect power $2^{2}$ out of $4 f$ .

$g = \pm \sqrt{\frac{1^{2} (n + 9 f)}{2^{2} f}}$

Rearrange the fraction $\frac{1^{2} (n + 9 f)}{2^{2} f}$ .

$g = \pm \sqrt{{(\frac{1}{2})}^{2} \frac{n + 9 f}{f}}$

Pull terms out from under the radical.

$g = \pm \frac{1}{2} \sqrt{\frac{n + 9 f}{f}}$

Rewrite $\sqrt{\frac{n + 9 f}{f}}$ as $\frac{\sqrt{n + 9 f}}{\sqrt{f}}$ .

$g = \pm \frac{1}{2} \cdot \frac{\sqrt{n + 9 f}}{\sqrt{f}}$

Multiply $\frac{\sqrt{n + 9 f}}{\sqrt{f}}$ by $\frac{\sqrt{f}}{\sqrt{f}}$ .

$g = \pm \frac{1}{2} (\frac{\sqrt{n + 9 f}}{\sqrt{f}} \cdot \frac{\sqrt{f}}{\sqrt{f}})$

Combine and simplify the denominator.

Multiply $\frac{\sqrt{n + 9 f}}{\sqrt{f}}$ by $\frac{\sqrt{f}}{\sqrt{f}}$ .

$g = \pm \frac{1}{2} \cdot \frac{\sqrt{n + 9 f} \sqrt{f}}{\sqrt{f} \sqrt{f}}$

Raise $\sqrt{f}$ to the power of $1$ .

$g = \pm \frac{1}{2} \cdot \frac{\sqrt{n + 9 f} \sqrt{f}}{{\sqrt{f}}^{1} \sqrt{f}}$

Raise $\sqrt{f}$ to the power of $1$ .

$g = \pm \frac{1}{2} \cdot \frac{\sqrt{n + 9 f} \sqrt{f}}{{\sqrt{f}}^{1} {\sqrt{f}}^{1}}$

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$g = \pm \frac{1}{2} \cdot \frac{\sqrt{n + 9 f} \sqrt{f}}{{\sqrt{f}}^{1 + 1}}$

Add $1$ and $1$ .

$g = \pm \frac{1}{2} \cdot \frac{\sqrt{n + 9 f} \sqrt{f}}{{\sqrt{f}}^{2}}$

Rewrite ${\sqrt{f}}^{2}$ as $f$ .

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

$g = \pm \frac{1}{2} \cdot \frac{\sqrt{n + 9 f} \sqrt{f}}{{(f^{\frac{1}{2}})}^{2}}$

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

$g = \pm \frac{1}{2} \cdot \frac{\sqrt{n + 9 f} \sqrt{f}}{f^{\frac{1}{2} \cdot 2}}$

Combine $\frac{1}{2}$ and $2$ .

$g = \pm \frac{1}{2} \cdot \frac{\sqrt{n + 9 f} \sqrt{f}}{f^{\frac{2}{2}}}$

Cancel the common factor of $2$ .

Cancel the common factor.

$g = \pm \frac{1}{2} \cdot \frac{\sqrt{n + 9 f} \sqrt{f}}{f^{\frac{2}{2}}}$

Rewrite the expression.

$g = \pm \frac{1}{2} \cdot \frac{\sqrt{n + 9 f} \sqrt{f}}{f^{1}}$

Simplify.

$g = \pm \frac{1}{2} \cdot \frac{\sqrt{n + 9 f} \sqrt{f}}{f}$

Combine using the product rule for radicals.

$g = \pm \frac{1}{2} \cdot \frac{\sqrt{(n + 9 f) f}}{f}$

Multiply $\frac{1}{2}$ by $\frac{\sqrt{(n + 9 f) f}}{f}$ .

$g = \pm \frac{\sqrt{(n + 9 f) f}}{2 f}$

Reorder factors in $\pm \frac{\sqrt{(n + 9 f) f}}{2 f}$ .

$g = \pm \frac{\sqrt{f (n + 9 f)}}{2 f}$

Step 6

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

First, use the positive value of the $\pm$ to find the first solution.

$g = \frac{\sqrt{f (n + 9 f)}}{2 f}$

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

$g = - \frac{\sqrt{f (n + 9 f)}}{2 f}$

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

$g = \frac{\sqrt{f (n + 9 f)}}{2 f}$

$g = - \frac{\sqrt{f (n + 9 f)}}{2 f}$

$g = \frac{\sqrt{f (n + 9 f)}}{2 f}$

$g = - \frac{\sqrt{f (n + 9 f)}}{2 f}$

Solve for g f(4g^2-9)=n

Related Questions

Support LCMGCF.com by disabling your adblocker.