Solve left(x^2-4x+2right)^2=x^2-4x+44

$${ \left( { x }^{ 2 } -4x+2 \right) }^{ 2 } = { x }^{ 2 } -4x+44$$

Solve for x (complex solution)

$$x=5$$

$$x=-1$$

$$x=2+2i$$

$$x=2-2i$$

Solution Steps

Square x^{2}-4x+2.

$$x^{4}-8x^{3}+20x^{2}-16x+4=x^{2}-4x+44$$

Subtract x^{2} from both sides.

$$x^{4}-8x^{3}+20x^{2}-16x+4-x^{2}=-4x+44$$

Combine 20x^{2} and -x^{2} to get 19x^{2}.

$$x^{4}-8x^{3}+19x^{2}-16x+4=-4x+44$$

Add 4x to both sides.

$$x^{4}-8x^{3}+19x^{2}-16x+4+4x=44$$

Combine -16x and 4x to get -12x.

$$x^{4}-8x^{3}+19x^{2}-12x+4=44$$

Subtract 44 from both sides.

$$x^{4}-8x^{3}+19x^{2}-12x+4-44=0$$

Subtract 44 from 4 to get -40.

$$x^{4}-8x^{3}+19x^{2}-12x-40=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -40 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±40,±20,±10,±8,±5,±4,±2,±1$$

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}-8x^{3}+19x^{2}-12x-40 by x+1 to get x^{3}-9x^{2}+28x-40. Solve the equation where the result equals to 0.

$$x^{3}-9x^{2}+28x-40=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -40 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±40,±20,±10,±8,±5,±4,±2,±1$$

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.

$$x=5$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{3}-9x^{2}+28x-40 by x-5 to get x^{2}-4x+8. Solve the equation where the result equals to 0.

$$x^{2}-4x+8=0$$

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, -4 for b, and 8 for c in the quadratic formula.

$$x=\frac{-\left(-4\right)±\sqrt{\left(-4\right)^{2}-4\times 1\times 8}}{2}$$

Do the calculations.

$$x=\frac{4±\sqrt{-16}}{2}$$

Solve the equation x^{2}-4x+8=0 when ± is plus and when ± is minus.

$$x=2-2i x=2+2i$$

List all found solutions.

$$x=-1 x=5 x=2-2i x=2+2i$$

Solve for x

$$x=-1$$

$$x=5$$

Solution Steps

Square x^{2}-4x+2.

$$x^{4}-8x^{3}+20x^{2}-16x+4=x^{2}-4x+44$$

Subtract x^{2} from both sides.

$$x^{4}-8x^{3}+20x^{2}-16x+4-x^{2}=-4x+44$$

Combine 20x^{2} and -x^{2} to get 19x^{2}.

$$x^{4}-8x^{3}+19x^{2}-16x+4=-4x+44$$

Add 4x to both sides.

$$x^{4}-8x^{3}+19x^{2}-16x+4+4x=44$$

Combine -16x and 4x to get -12x.

$$x^{4}-8x^{3}+19x^{2}-12x+4=44$$

Subtract 44 from both sides.

$$x^{4}-8x^{3}+19x^{2}-12x+4-44=0$$

Subtract 44 from 4 to get -40.

$$x^{4}-8x^{3}+19x^{2}-12x-40=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -40 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±40,±20,±10,±8,±5,±4,±2,±1$$

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}-8x^{3}+19x^{2}-12x-40 by x+1 to get x^{3}-9x^{2}+28x-40. Solve the equation where the result equals to 0.

$$x^{3}-9x^{2}+28x-40=0$$

By Rational Root Theorem, all rational roots of a polynomial are in the form \frac{p}{q}, where p divides the constant term -40 and q divides the leading coefficient 1. List all candidates \frac{p}{q}.

$$±40,±20,±10,±8,±5,±4,±2,±1$$

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.

$$x=5$$

By Factor theorem, x-k is a factor of the polynomial for each root k. Divide x^{3}-9x^{2}+28x-40 by x-5 to get x^{2}-4x+8. Solve the equation where the result equals to 0.

$$x^{2}-4x+8=0$$

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, -4 for b, and 8 for c in the quadratic formula.

$$x=\frac{-\left(-4\right)±\sqrt{\left(-4\right)^{2}-4\times 1\times 8}}{2}$$

Do the calculations.

$$x=\frac{4±\sqrt{-16}}{2}$$

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.

$$x=-1 x=5$$