Solve n^2+3/n+2-1=1

$$n ^ { 2 } + \frac { 3 } { n + 2 } - 1 = 1$$

Solve for n

$$n=\frac{-\sqrt{5}-3}{2}\approx -2.618033989$$

$$n=1$$

$$n=\frac{\sqrt{5}-3}{2}\approx -0.381966011$$

Solution Steps

Variable n cannot be equal to -2 since division by zero is not defined. Multiply both sides of the equation by n+2.

$$\left(n+2\right)n^{2}+3+\left(n+2\right)\left(-1\right)=n+2$$

Use the distributive property to multiply n+2 by n^{2}.

$$n^{3}+2n^{2}+3+\left(n+2\right)\left(-1\right)=n+2$$

Use the distributive property to multiply n+2 by -1.

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

Subtract 2 from 3 to get 1.

$$n^{3}+2n^{2}+1-n=n+2$$

Subtract n from both sides.

$$n^{3}+2n^{2}+1-n-n=2$$

Combine -n and -n to get -2n.

$$n^{3}+2n^{2}+1-2n=2$$

Subtract 2 from both sides.

$$n^{3}+2n^{2}+1-2n-2=0$$

Subtract 2 from 1 to get -1.

$$n^{3}+2n^{2}-1-2n=0$$

Rearrange the equation to put it in standard form. Place the terms in order from highest to lowest power.

$$n^{3}+2n^{2}-2n-1=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -1 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±1$$

Find one such root by trying out all the integer values, starting from the smallest by absolute value. If no integer roots are found, try out fractions.

$$n=1$$

By Factor theorem, n-k is a factor of the polynomial for each root k. Divide n^{3}+2n^{2}-2n-1 by n-1 to get n^{2}+3n+1. Solve the equation where the result equals to 0.

$$n^{2}+3n+1=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Substitute 1 for a, 3 for b, and 1 for c in the quadratic formula.

$$n=\frac{-3±\sqrt{3^{2}-4\times 1\times 1}}{2}$$

Do the calculations.

$$n=\frac{-3±\sqrt{5}}{2}$$

Solve the equation n^{2}+3n+1=0 when ± is plus and when ± is minus.

$$n=\frac{-\sqrt{5}-3}{2} n=\frac{\sqrt{5}-3}{2}$$

List all found solutions.

$$n=1 n=\frac{-\sqrt{5}-3}{2} n=\frac{\sqrt{5}-3}{2}$$