Solve x^2+2x-6/x^2+2x=1

$$x ^ { 2 } + 2 x - \frac { 6 } { x ^ { 2 } + 2 x } = 1$$

Solve for x (complex solution)

$$x=-1-i$$

$$x=-1+i$$

$$x=-3$$

$$x=1$$

Solution Steps

Variable x cannot be equal to any of the values -2,0 since division by zero is not defined. Multiply both sides of the equation by x\left(x+2\right).

$$x\left(x+2\right)x^{2}+2xx\left(x+2\right)-6=x\left(x+2\right)$$

Use the distributive property to multiply x by x+2.

$$\left(x^{2}+2x\right)x^{2}+2xx\left(x+2\right)-6=x\left(x+2\right)$$

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

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

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

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

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

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

Combine 2x^{3} and 2x^{3} to get 4x^{3}.

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

Use the distributive property to multiply x by x+2.

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

Subtract x^{2} from both sides.

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

Combine 4x^{2} and -x^{2} to get 3x^{2}.

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

Subtract 2x from both sides.

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

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

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

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -6 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±6,±3,±2,±1$$

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}+4x^{3}+3x^{2}-2x-6 by x-1 to get x^{3}+5x^{2}+8x+6. Solve the equation where the result equals to 0.

$$x^{3}+5x^{2}+8x+6=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 6 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±6,±3,±2,±1$$

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.

$$x=-3$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{3}+5x^{2}+8x+6 by x+3 to get x^{2}+2x+2. Solve the equation where the result equals to 0.

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

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

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

Do the calculations.

$$x=\frac{-2±\sqrt{-4}}{2}$$

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

$$x=-1-i x=-1+i$$

List all found solutions.

$$x=1 x=-3 x=-1-i x=-1+i$$

Solve for x

$$x=1$$

$$x=-3$$

Solution Steps

Variable x cannot be equal to any of the values -2,0 since division by zero is not defined. Multiply both sides of the equation by x\left(x+2\right).

$$x\left(x+2\right)x^{2}+2xx\left(x+2\right)-6=x\left(x+2\right)$$

Use the distributive property to multiply x by x+2.

$$\left(x^{2}+2x\right)x^{2}+2xx\left(x+2\right)-6=x\left(x+2\right)$$

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

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

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

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

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

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

Combine 2x^{3} and 2x^{3} to get 4x^{3}.

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

Use the distributive property to multiply x by x+2.

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

Subtract x^{2} from both sides.

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

Combine 4x^{2} and -x^{2} to get 3x^{2}.

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

Subtract 2x from both sides.

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

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

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

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -6 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±6,±3,±2,±1$$

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}+4x^{3}+3x^{2}-2x-6 by x-1 to get x^{3}+5x^{2}+8x+6. Solve the equation where the result equals to 0.

$$x^{3}+5x^{2}+8x+6=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 6 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±6,±3,±2,±1$$

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.

$$x=-3$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{3}+5x^{2}+8x+6 by x+3 to get x^{2}+2x+2. Solve the equation where the result equals to 0.

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

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

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

Do the calculations.

$$x=\frac{-2±\sqrt{-4}}{2}$$

Since the square root of a negative number is not defined in the real field, there are no solutions.

$$x\in \emptyset$$

List all found solutions.

$$x=1 x=-3$$