Solve x^2+3x-20/x^2+3x=8

$$x ^ { 2 } + 3 x - \frac { 20 } { x ^ { 2 } + 3 x } = 8$$

Solve for x

$$x=-2$$

$$x=-1$$

$$x=2$$

$$x=-5$$

Solution Steps

Variable x cannot be equal to any of the values -3,0 since division by zero is not defined. Multiply both sides of the equation by x\left(x+3\right).

$$x\left(x+3\right)x^{2}+3xx\left(x+3\right)-20=8x\left(x+3\right)$$

Use the distributive property to multiply x by x+3.

$$\left(x^{2}+3x\right)x^{2}+3xx\left(x+3\right)-20=8x\left(x+3\right)$$

Use the distributive property to multiply x^{2}+3x by x^{2}.

$$x^{4}+3x^{3}+3xx\left(x+3\right)-20=8x\left(x+3\right)$$

Multiply x and x to get x^{2}.

$$x^{4}+3x^{3}+3x^{2}\left(x+3\right)-20=8x\left(x+3\right)$$

Use the distributive property to multiply 3x^{2} by x+3.

$$x^{4}+3x^{3}+3x^{3}+9x^{2}-20=8x\left(x+3\right)$$

Combine 3x^{3} and 3x^{3} to get 6x^{3}.

$$x^{4}+6x^{3}+9x^{2}-20=8x\left(x+3\right)$$

Use the distributive property to multiply 8x by x+3.

$$x^{4}+6x^{3}+9x^{2}-20=8x^{2}+24x$$

Subtract 8x^{2} from both sides.

$$x^{4}+6x^{3}+9x^{2}-20-8x^{2}=24x$$

Combine 9x^{2} and -8x^{2} to get x^{2}.

$$x^{4}+6x^{3}+x^{2}-20=24x$$

Subtract 24x from both sides.

$$x^{4}+6x^{3}+x^{2}-20-24x=0$$

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

$$x^{4}+6x^{3}+x^{2}-24x-20=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -20 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±20,±10,±5,±4,±2,±1$$

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.

$$x=-1$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{4}+6x^{3}+x^{2}-24x-20 by x+1 to get x^{3}+5x^{2}-4x-20. Solve the equation where the result equals to 0.

$$x^{3}+5x^{2}-4x-20=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -20 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±20,±10,±5,±4,±2,±1$$

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}+5x^{2}-4x-20 by x-2 to get x^{2}+7x+10. Solve the equation where the result equals to 0.

$$x^{2}+7x+10=0$$

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, 7 for b, and 10 for c in the quadratic formula.

$$x=\frac{-7±\sqrt{7^{2}-4\times 1\times 10}}{2}$$

Do the calculations.

$$x=\frac{-7±3}{2}$$

Solve the equation x^{2}+7x+10=0 when ± is plus and when ± is minus.

$$x=-5 x=-2$$

List all found solutions.

$$x=-1 x=2 x=-5 x=-2$$