Solve x^4=(2x-3)^2

$$x ^ { 4 } = ( 2 x - 3 ) ^ { 2 }$$

Solve for x (complex solution)

$$x=-3$$

$$x=1$$

$$x=-\sqrt{2}i+1\approx 1-1.414213562i$$

$$x=1+\sqrt{2}i\approx 1+1.414213562i$$

Solution Steps

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

$$x^{4}=4x^{2}-12x+9$$

Subtract 4x^{2} from both sides.

$$x^{4}-4x^{2}=-12x+9$$

Add 12x to both sides.

$$x^{4}-4x^{2}+12x=9$$

Subtract 9 from both sides.

$$x^{4}-4x^{2}+12x-9=0$$

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}-4x^{2}+12x-9 by x-1 to get x^{3}+x^{2}-3x+9. Solve the equation where the result equals to 0.

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

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

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

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, -2 for b, and 3 for c in the quadratic formula.

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

Do the calculations.

$$x=\frac{2±\sqrt{-8}}{2}$$

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

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

List all found solutions.

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

Solve for x

$$x=-3$$

$$x=1$$

Solution Steps

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

$$x^{4}=4x^{2}-12x+9$$

Subtract 4x^{2} from both sides.

$$x^{4}-4x^{2}=-12x+9$$

Add 12x to both sides.

$$x^{4}-4x^{2}+12x=9$$

Subtract 9 from both sides.

$$x^{4}-4x^{2}+12x-9=0$$

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}-4x^{2}+12x-9 by x-1 to get x^{3}+x^{2}-3x+9. Solve the equation where the result equals to 0.

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

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

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

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, -2 for b, and 3 for c in the quadratic formula.

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

Do the calculations.

$$x=\frac{2±\sqrt{-8}}{2}$$

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.

$$x=1 x=-3$$