Solve x^2+x+1=42/x^2+x

$$x ^ { 2 } + x + 1 = \frac { 42 } { x ^ { 2 } + x }$$

Solve for x (complex solution)

$$x=\frac{-3\sqrt{3}i-1}{2}\approx -0.5-2.598076211i$$

$$x=\frac{-1+3\sqrt{3}i}{2}\approx -0.5+2.598076211i$$

$$x=-3$$

$$x=2$$

Solution Steps

Variable x cannot be equal to any of the values -1,0 since division by zero is not defined. Multiply both sides of the equation by x\left(x+1\right).

$$x\left(x+1\right)x^{2}+x\left(x+1\right)x+x\left(x+1\right)=42$$

Use the distributive property to multiply x by x+1.

$$\left(x^{2}+x\right)x^{2}+x\left(x+1\right)x+x\left(x+1\right)=42$$

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

$$x^{4}+x^{3}+x\left(x+1\right)x+x\left(x+1\right)=42$$

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

$$x^{4}+x^{3}+x^{2}\left(x+1\right)+x\left(x+1\right)=42$$

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

$$x^{4}+x^{3}+x^{3}+x^{2}+x\left(x+1\right)=42$$

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

$$x^{4}+2x^{3}+x^{2}+x\left(x+1\right)=42$$

Use the distributive property to multiply x by x+1.

$$x^{4}+2x^{3}+x^{2}+x^{2}+x=42$$

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

$$x^{4}+2x^{3}+2x^{2}+x=42$$

Subtract 42 from both sides.

$$x^{4}+2x^{3}+2x^{2}+x-42=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -42 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±42,±21,±14,±7,±6,±3,±2,±1$$

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.

$$x=2$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{4}+2x^{3}+2x^{2}+x-42 by x-2 to get x^{3}+4x^{2}+10x+21. Solve the equation where the result equals to 0.

$$x^{3}+4x^{2}+10x+21=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 21 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±21,±7,±3,±1$$

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}+4x^{2}+10x+21 by x+3 to get x^{2}+x+7. Solve the equation where the result equals to 0.

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

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

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

Do the calculations.

$$x=\frac{-1±\sqrt{-27}}{2}$$

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

$$x=\frac{-3i\sqrt{3}-1}{2} x=\frac{-1+3i\sqrt{3}}{2}$$

List all found solutions.

$$x=2 x=-3 x=\frac{-3i\sqrt{3}-1}{2} x=\frac{-1+3i\sqrt{3}}{2}$$

Solve for x

$$x=2$$

$$x=-3$$

Solution Steps

Variable x cannot be equal to any of the values -1,0 since division by zero is not defined. Multiply both sides of the equation by x\left(x+1\right).

$$x\left(x+1\right)x^{2}+x\left(x+1\right)x+x\left(x+1\right)=42$$

Use the distributive property to multiply x by x+1.

$$\left(x^{2}+x\right)x^{2}+x\left(x+1\right)x+x\left(x+1\right)=42$$

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

$$x^{4}+x^{3}+x\left(x+1\right)x+x\left(x+1\right)=42$$

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

$$x^{4}+x^{3}+x^{2}\left(x+1\right)+x\left(x+1\right)=42$$

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

$$x^{4}+x^{3}+x^{3}+x^{2}+x\left(x+1\right)=42$$

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

$$x^{4}+2x^{3}+x^{2}+x\left(x+1\right)=42$$

Use the distributive property to multiply x by x+1.

$$x^{4}+2x^{3}+x^{2}+x^{2}+x=42$$

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

$$x^{4}+2x^{3}+2x^{2}+x=42$$

Subtract 42 from both sides.

$$x^{4}+2x^{3}+2x^{2}+x-42=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -42 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±42,±21,±14,±7,±6,±3,±2,±1$$

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.

$$x=2$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{4}+2x^{3}+2x^{2}+x-42 by x-2 to get x^{3}+4x^{2}+10x+21. Solve the equation where the result equals to 0.

$$x^{3}+4x^{2}+10x+21=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term 21 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±21,±7,±3,±1$$

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}+4x^{2}+10x+21 by x+3 to get x^{2}+x+7. Solve the equation where the result equals to 0.

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

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

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

Do the calculations.

$$x=\frac{-1±\sqrt{-27}}{2}$$

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.

$$x=2 x=-3$$