Solve x^2(x-1)^2=(x+3)^2

$$x ^ { 2 } ( x - 1 ) ^ { 2 } = ( x + 3 ) ^ { 2 }$$

Solve for x (complex solution)

$$x=-1$$

$$x=3$$

$$x=\sqrt{3}i\approx 1.732050808i$$

$$x=-\sqrt{3}i\approx -0-1.732050808i$$

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-1\right)^{2}.

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

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

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

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+3\right)^{2}.

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

Subtract x^{2} from both sides.

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

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

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

Subtract 6x from both sides.

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

Subtract 9 from both sides.

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

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.

$$x=-1$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{4}-2x^{3}-6x-9 by x+1 to get x^{3}-3x^{2}+3x-9. Solve the equation where the result equals to 0.

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

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

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

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, 0 for b, and 3 for c in the quadratic formula.

$$x=\frac{0±\sqrt{0^{2}-4\times 1\times 3}}{2}$$

Do the calculations.

$$x=\frac{0±\sqrt{-12}}{2}$$

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

$$x=-\sqrt{3}i x=\sqrt{3}i$$

List all found solutions.

$$x=-1 x=3 x=-\sqrt{3}i x=\sqrt{3}i$$

Solve for x

$$x=-1$$

$$x=3$$

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-1\right)^{2}.

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

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

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

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+3\right)^{2}.

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

Subtract x^{2} from both sides.

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

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

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

Subtract 6x from both sides.

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

Subtract 9 from both sides.

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

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.

$$x=-1$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{4}-2x^{3}-6x-9 by x+1 to get x^{3}-3x^{2}+3x-9. Solve the equation where the result equals to 0.

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

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

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

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, 0 for b, and 3 for c in the quadratic formula.

$$x=\frac{0±\sqrt{0^{2}-4\times 1\times 3}}{2}$$

Do the calculations.

$$x=\frac{0±\sqrt{-12}}{2}$$

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.

$$x=-1 x=3$$