Solve x^2-x*2x+1-x^2=2x^2+4x-x-1

$${ x }^{ 2 } -x \times 2x+1- { x }^{ 2 } =2 { x }^{ 2 } +4x-x-1$$

Solve for x

$$x=\frac{\sqrt{41}-3}{8}\approx 0.42539053$$

$$x=\frac{-\sqrt{41}-3}{8}\approx -1.17539053$$

Steps Using the Quadratic Formula

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

$$x^{2}-x^{2}\times 2+1-x^{2}=2x^{2}+4x-x-1$$

Combine x^{2} and -x^{2}\times 2 to get -x^{2}.

$$-x^{2}+1-x^{2}=2x^{2}+4x-x-1$$

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

$$-2x^{2}+1=2x^{2}+4x-x-1$$

Combine 4x and -x to get 3x.

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

Subtract 2x^{2} from both sides.

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

Combine -2x^{2} and -2x^{2} to get -4x^{2}.

$$-4x^{2}+1=3x-1$$

Subtract 3x from both sides.

$$-4x^{2}+1-3x=-1$$

Add 1 to both sides.

$$-4x^{2}+1-3x+1=0$$

Add 1 and 1 to get 2.

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

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

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

This equation is in standard form: ax^{2}+bx+c=0. Substitute -4 for a, -3 for b, and 2 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Square -3.

$$x=\frac{-\left(-3\right)±\sqrt{9-4\left(-4\right)\times 2}}{2\left(-4\right)}$$

Multiply -4 times -4.

$$x=\frac{-\left(-3\right)±\sqrt{9+16\times 2}}{2\left(-4\right)}$$

Multiply 16 times 2.

$$x=\frac{-\left(-3\right)±\sqrt{9+32}}{2\left(-4\right)}$$

Add 9 to 32.

$$x=\frac{-\left(-3\right)±\sqrt{41}}{2\left(-4\right)}$$

The opposite of -3 is 3.

$$x=\frac{3±\sqrt{41}}{2\left(-4\right)}$$

Multiply 2 times -4.

$$x=\frac{3±\sqrt{41}}{-8}$$

Now solve the equation x=\frac{3±\sqrt{41}}{-8} when ± is plus. Add 3 to \sqrt{41}.

$$x=\frac{\sqrt{41}+3}{-8}$$

Divide 3+\sqrt{41} by -8.

$$x=\frac{-\sqrt{41}-3}{8}$$

Now solve the equation x=\frac{3±\sqrt{41}}{-8} when ± is minus. Subtract \sqrt{41} from 3.

$$x=\frac{3-\sqrt{41}}{-8}$$

Divide 3-\sqrt{41} by -8.

$$x=\frac{\sqrt{41}-3}{8}$$

The equation is now solved.

$$x=\frac{-\sqrt{41}-3}{8} x=\frac{\sqrt{41}-3}{8}$$

Steps for Completing the Square

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

$$x^{2}-x^{2}\times 2+1-x^{2}=2x^{2}+4x-x-1$$

Combine x^{2} and -x^{2}\times 2 to get -x^{2}.

$$-x^{2}+1-x^{2}=2x^{2}+4x-x-1$$

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

$$-2x^{2}+1=2x^{2}+4x-x-1$$

Combine 4x and -x to get 3x.

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

Subtract 2x^{2} from both sides.

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

Combine -2x^{2} and -2x^{2} to get -4x^{2}.

$$-4x^{2}+1=3x-1$$

Subtract 3x from both sides.

$$-4x^{2}+1-3x=-1$$

Subtract 1 from both sides.

$$-4x^{2}-3x=-1-1$$

Subtract 1 from -1 to get -2.

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

Divide both sides by -4.

$$\frac{-4x^{2}-3x}{-4}=-\frac{2}{-4}$$

Dividing by -4 undoes the multiplication by -4.

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

Divide -3 by -4.

$$x^{2}+\frac{3}{4}x=-\frac{2}{-4}$$

Reduce the fraction \frac{-2}{-4} to lowest terms by extracting and canceling out 2.

$$x^{2}+\frac{3}{4}x=\frac{1}{2}$$

Divide \frac{3}{4}, the coefficient of the x term, by 2 to get \frac{3}{8}. Then add the square of \frac{3}{8} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{3}{4}x+\left(\frac{3}{8}\right)^{2}=\frac{1}{2}+\left(\frac{3}{8}\right)^{2}$$

Square \frac{3}{8} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{3}{4}x+\frac{9}{64}=\frac{1}{2}+\frac{9}{64}$$

Add \frac{1}{2} to \frac{9}{64} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{3}{4}x+\frac{9}{64}=\frac{41}{64}$$

Factor x^{2}+\frac{3}{4}x+\frac{9}{64}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{3}{8}\right)^{2}=\frac{41}{64}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{3}{8}\right)^{2}}=\sqrt{\frac{41}{64}}$$

Simplify.

$$x+\frac{3}{8}=\frac{\sqrt{41}}{8} x+\frac{3}{8}=-\frac{\sqrt{41}}{8}$$

Subtract \frac{3}{8} from both sides of the equation.

$$x=\frac{\sqrt{41}-3}{8} x=\frac{-\sqrt{41}-3}{8}$$