Solve 3/n^2=n-4/3n^2+2/3n^022

$$\frac{ 3 }{ { n }^{ 2 } } = \frac{ n-4 }{ 3 { n }^{ 2 } } + \frac{ 2 }{ 3 { n }^{ 022 } }$$

Solve for n

$$n = \frac{\sqrt{105} - 1}{4} \approx 2.311737691$$

$$n=\frac{-\sqrt{105}-1}{4}\approx -2.811737691$$

Steps Using the Quadratic Formula

Variable n cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by 3n^{2}, the least common multiple of n^{2},3n^{2}.

$$3\times 3=n-4+n^{2}\times 2$$

Multiply 3 and 3 to get 9.

$$9=n-4+n^{2}\times 2$$

Swap sides so that all variable terms are on the left hand side.

$$n-4+n^{2}\times 2=9$$

Subtract 9 from both sides.

$$n-4+n^{2}\times 2-9=0$$

Subtract 9 from -4 to get -13.

$$n-13+n^{2}\times 2=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$2n^{2}+n-13=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, 1 for b, and -13 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$n=\frac{-1±\sqrt{1^{2}-4\times 2\left(-13\right)}}{2\times 2}$$

Square 1.

$$n=\frac{-1±\sqrt{1-4\times 2\left(-13\right)}}{2\times 2}$$

Multiply -4 times 2.

$$n=\frac{-1±\sqrt{1-8\left(-13\right)}}{2\times 2}$$

Multiply -8 times -13.

$$n=\frac{-1±\sqrt{1+104}}{2\times 2}$$

Add 1 to 104.

$$n=\frac{-1±\sqrt{105}}{2\times 2}$$

Multiply 2 times 2.

$$n=\frac{-1±\sqrt{105}}{4}$$

Now solve the equation n=\frac{-1±\sqrt{105}}{4} when ± is plus. Add -1 to \sqrt{105}.

$$n=\frac{\sqrt{105}-1}{4}$$

Now solve the equation n=\frac{-1±\sqrt{105}}{4} when ± is minus. Subtract \sqrt{105} from -1.

$$n=\frac{-\sqrt{105}-1}{4}$$

The equation is now solved.

$$n=\frac{\sqrt{105}-1}{4} n=\frac{-\sqrt{105}-1}{4}$$

Steps for Completing the Square

Variable n cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by 3n^{2}, the least common multiple of n^{2},3n^{2}.

$$3\times 3=n-4+n^{2}\times 2$$

Multiply 3 and 3 to get 9.

$$9=n-4+n^{2}\times 2$$

Swap sides so that all variable terms are on the left hand side.

$$n-4+n^{2}\times 2=9$$

Add 4 to both sides.

$$n+n^{2}\times 2=9+4$$

Add 9 and 4 to get 13.

$$n+n^{2}\times 2=13$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$2n^{2}+n=13$$

Divide both sides by 2.

$$\frac{2n^{2}+n}{2}=\frac{13}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$n^{2}+\frac{1}{2}n=\frac{13}{2}$$

Divide \frac{1}{2}, the coefficient of the x term, by 2 to get \frac{1}{4}. Then add the square of \frac{1}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$n^{2}+\frac{1}{2}n+\left(\frac{1}{4}\right)^{2}=\frac{13}{2}+\left(\frac{1}{4}\right)^{2}$$

Square \frac{1}{4} by squaring both the numerator and the denominator of the fraction.

$$n^{2}+\frac{1}{2}n+\frac{1}{16}=\frac{13}{2}+\frac{1}{16}$$

Add \frac{13}{2} to \frac{1}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$n^{2}+\frac{1}{2}n+\frac{1}{16}=\frac{105}{16}$$

Factor n^{2}+\frac{1}{2}n+\frac{1}{16}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(n+\frac{1}{4}\right)^{2}=\frac{105}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(n+\frac{1}{4}\right)^{2}}=\sqrt{\frac{105}{16}}$$

Simplify.

$$n+\frac{1}{4}=\frac{\sqrt{105}}{4} n+\frac{1}{4}=-\frac{\sqrt{105}}{4}$$

Subtract \frac{1}{4} from both sides of the equation.

$$n=\frac{\sqrt{105}-1}{4} n=\frac{-\sqrt{105}-1}{4}$$