Solve (x^2-x)^2-3(x^2-x)=18

$$( x ^ { 2 } - x ) ^ { 2 } - 3 ( x ^ { 2 } - x ) = 18$$

Solve for x (complex solution)

$$x=3$$

$$x=-2$$

$$x=\frac{1+\sqrt{11}i}{2}\approx 0.5+1.658312395i$$

$$x=\frac{-\sqrt{11}i+1}{2}\approx 0.5-1.658312395i$$

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x^{2}-x\right)^{2}.

$$\left(x^{2}\right)^{2}-2x^{2}x+x^{2}-3\left(x^{2}-x\right)=18$$

To raise a power to another power, multiply the exponents. Multiply 2 and 2 to get 4.

$$x^{4}-2x^{2}x+x^{2}-3\left(x^{2}-x\right)=18$$

To multiply powers of the same base, add their exponents. Add 2 and 1 to get 3.

$$x^{4}-2x^{3}+x^{2}-3\left(x^{2}-x\right)=18$$

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

$$x^{4}-2x^{3}+x^{2}-3x^{2}+3x=18$$

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

$$x^{4}-2x^{3}-2x^{2}+3x=18$$

Subtract 18 from both sides.

$$x^{4}-2x^{3}-2x^{2}+3x-18=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -18 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±18,±9,±6,±3,±2,±1$$

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}+3x-18 by x+2 to get x^{3}-4x^{2}+6x-9. Solve the equation where the result equals to 0.

$$x^{3}-4x^{2}+6x-9=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -9 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±9,±3,±1$$

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}+6x-9 by x-3 to get x^{2}-x+3. Solve the equation where the result equals to 0.

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

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 3 for c in the quadratic formula.

$$x=\frac{-\left(-1\right)±\sqrt{\left(-1\right)^{2}-4\times 1\times 3}}{2}$$

Do the calculations.

$$x=\frac{1±\sqrt{-11}}{2}$$

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

$$x=\frac{-\sqrt{11}i+1}{2} x=\frac{1+\sqrt{11}i}{2}$$

List all found solutions.

$$x=-2 x=3 x=\frac{-\sqrt{11}i+1}{2} x=\frac{1+\sqrt{11}i}{2}$$

Solve for x

$$x=-2$$

$$x=3$$

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x^{2}-x\right)^{2}.

$$\left(x^{2}\right)^{2}-2x^{2}x+x^{2}-3\left(x^{2}-x\right)=18$$

To raise a power to another power, multiply the exponents. Multiply 2 and 2 to get 4.

$$x^{4}-2x^{2}x+x^{2}-3\left(x^{2}-x\right)=18$$

To multiply powers of the same base, add their exponents. Add 2 and 1 to get 3.

$$x^{4}-2x^{3}+x^{2}-3\left(x^{2}-x\right)=18$$

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

$$x^{4}-2x^{3}+x^{2}-3x^{2}+3x=18$$

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

$$x^{4}-2x^{3}-2x^{2}+3x=18$$

Subtract 18 from both sides.

$$x^{4}-2x^{3}-2x^{2}+3x-18=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -18 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±18,±9,±6,±3,±2,±1$$

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}+3x-18 by x+2 to get x^{3}-4x^{2}+6x-9. Solve the equation where the result equals to 0.

$$x^{3}-4x^{2}+6x-9=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -9 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±9,±3,±1$$

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}+6x-9 by x-3 to get x^{2}-x+3. Solve the equation where the result equals to 0.

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

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 3 for c in the quadratic formula.

$$x=\frac{-\left(-1\right)±\sqrt{\left(-1\right)^{2}-4\times 1\times 3}}{2}$$

Do the calculations.

$$x=\frac{1±\sqrt{-11}}{2}$$

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.

$$x=-2 x=3$$