Solve 1=x+8/x^2+x-2-x+4/x^2+3x+2

$$1 = \frac { x + 8 } { x ^ { 2 } + x - 2 } - \frac { x + 4 } { x ^ { 2 } + 3 x + 2 }$$

Solve for x

$$x=\sqrt{7}\approx 2.645751311$$

$$x=-\sqrt{7}\approx -2.645751311$$

Solution Steps

Variable x cannot be equal to any of the values -2,-1,1 since division by zero is not defined. Multiply both sides of the equation by \left(x-1\right)\left(x+1\right)\left(x+2\right), the least common multiple of x^{2}+x-2,x^{2}+3x+2.

$$\left(x-1\right)\left(x+1\right)\left(x+2\right)=\left(x+1\right)\left(x+8\right)-\left(x-1\right)\left(x+4\right)$$

Use the distributive property to multiply x-1 by x+1 and combine like terms.

$$\left(x^{2}-1\right)\left(x+2\right)=\left(x+1\right)\left(x+8\right)-\left(x-1\right)\left(x+4\right)$$

Use the distributive property to multiply x^{2}-1 by x+2.

$$x^{3}+2x^{2}-x-2=\left(x+1\right)\left(x+8\right)-\left(x-1\right)\left(x+4\right)$$

Use the distributive property to multiply x+1 by x+8 and combine like terms.

$$x^{3}+2x^{2}-x-2=x^{2}+9x+8-\left(x-1\right)\left(x+4\right)$$

Use the distributive property to multiply x-1 by x+4 and combine like terms.

$$x^{3}+2x^{2}-x-2=x^{2}+9x+8-\left(x^{2}+3x-4\right)$$

To find the opposite of x^{2}+3x-4, find the opposite of each term.

$$x^{3}+2x^{2}-x-2=x^{2}+9x+8-x^{2}-3x+4$$

Combine x^{2} and -x^{2} to get 0.

$$x^{3}+2x^{2}-x-2=9x+8-3x+4$$

Combine 9x and -3x to get 6x.

$$x^{3}+2x^{2}-x-2=6x+8+4$$

Add 8 and 4 to get 12.

$$x^{3}+2x^{2}-x-2=6x+12$$

Subtract 6x from both sides.

$$x^{3}+2x^{2}-x-2-6x=12$$

Combine -x and -6x to get -7x.

$$x^{3}+2x^{2}-7x-2=12$$

Subtract 12 from both sides.

$$x^{3}+2x^{2}-7x-2-12=0$$

Subtract 12 from -2 to get -14.

$$x^{3}+2x^{2}-7x-14=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -14 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±14,±7,±2,±1$$

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.

$$x=-2$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{3}+2x^{2}-7x-14 by x+2 to get x^{2}-7. Solve the equation where the result equals to 0.

$$x^{2}-7=0$$

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, 0 for b, and -7 for c in the quadratic formula.

$$x=\frac{0±\sqrt{0^{2}-4\times 1\left(-7\right)}}{2}$$

Do the calculations.

$$x=\frac{0±2\sqrt{7}}{2}$$

Solve the equation x^{2}-7=0 when ± is plus and when ± is minus.

$$x=-\sqrt{7} x=\sqrt{7}$$

Remove the values that the variable cannot be equal to.

$$x\in \emptyset$$

List all found solutions.

$$x=-2 x=-\sqrt{7} x=\sqrt{7}$$

Variable x cannot be equal to -2.

$$x=\sqrt{7} x=-\sqrt{7}$$