Solve x^2-2x^2-2x+5=4

$$x ^ { 2 } - 2 x ^ { 2 } - 2 x + 5 = 4$$

Solve for x (complex solution)

$$x=\sqrt{2}-1\approx 0.414213562$$

$$x=-\left(\sqrt{2}+1\right)\approx -2.414213562$$

Steps Using the Quadratic Formula

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

$$-x^{2}-2x+5=4$$

Subtract 4 from both sides.

$$-x^{2}-2x+5-4=0$$

Subtract 4 from 5 to get 1.

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

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, -2 for b, and 1 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Square -2.

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

Multiply -4 times -1.

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

Add 4 to 4.

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

Take the square root of 8.

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

The opposite of -2 is 2.

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

Multiply 2 times -1.

$$x=\frac{2±2\sqrt{2}}{-2}$$

Now solve the equation x=\frac{2±2\sqrt{2}}{-2} when ± is plus. Add 2 to 2\sqrt{2}.

$$x=\frac{2\sqrt{2}+2}{-2}$$

Divide 2+2\sqrt{2} by -2.

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

Now solve the equation x=\frac{2±2\sqrt{2}}{-2} when ± is minus. Subtract 2\sqrt{2} from 2.

$$x=\frac{2-2\sqrt{2}}{-2}$$

Divide 2-2\sqrt{2} by -2.

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

The equation is now solved.

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

Steps for Completing the Square

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

$$-x^{2}-2x+5=4$$

Subtract 5 from both sides.

$$-x^{2}-2x=4-5$$

Subtract 5 from 4 to get -1.

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

Divide both sides by -1.

$$\frac{-x^{2}-2x}{-1}=-\frac{1}{-1}$$

Dividing by -1 undoes the multiplication by -1.

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

Divide -2 by -1.

$$x^{2}+2x=-\frac{1}{-1}$$

Divide -1 by -1.

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

Divide 2, the coefficient of the x term, by 2 to get 1. Then add the square of 1 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square 1.

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

Add 1 to 1.

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

Factor x^{2}+2x+1. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

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

Take the square root of both sides of the equation.

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

Simplify.

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

Subtract 1 from both sides of the equation.

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

Solve for x

$$x=\sqrt{2}-1\approx 0.414213562$$

$$x=-\sqrt{2}-1\approx -2.414213562$$

Steps Using the Quadratic Formula

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

$$-x^{2}-2x+5=4$$

Subtract 4 from both sides.

$$-x^{2}-2x+5-4=0$$

Subtract 4 from 5 to get 1.

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

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, -2 for b, and 1 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Square -2.

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

Multiply -4 times -1.

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

Add 4 to 4.

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

Take the square root of 8.

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

The opposite of -2 is 2.

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

Multiply 2 times -1.

$$x=\frac{2±2\sqrt{2}}{-2}$$

Now solve the equation x=\frac{2±2\sqrt{2}}{-2} when ± is plus. Add 2 to 2\sqrt{2}.

$$x=\frac{2\sqrt{2}+2}{-2}$$

Divide 2+2\sqrt{2} by -2.

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

Now solve the equation x=\frac{2±2\sqrt{2}}{-2} when ± is minus. Subtract 2\sqrt{2} from 2.

$$x=\frac{2-2\sqrt{2}}{-2}$$

Divide 2-2\sqrt{2} by -2.

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

The equation is now solved.

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

Steps for Completing the Square

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

$$-x^{2}-2x+5=4$$

Subtract 5 from both sides.

$$-x^{2}-2x=4-5$$

Subtract 5 from 4 to get -1.

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

Divide both sides by -1.

$$\frac{-x^{2}-2x}{-1}=-\frac{1}{-1}$$

Dividing by -1 undoes the multiplication by -1.

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

Divide -2 by -1.

$$x^{2}+2x=-\frac{1}{-1}$$

Divide -1 by -1.

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

Divide 2, the coefficient of the x term, by 2 to get 1. Then add the square of 1 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square 1.

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

Add 1 to 1.

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

Factor x^{2}+2x+1. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

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

Take the square root of both sides of the equation.

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

Simplify.

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

Subtract 1 from both sides of the equation.

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