Solve x^2-1/2x=1-x

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

Solve for x

$$x=\frac{\sqrt{17}-1}{4}\approx 0.780776406$$

$$x=\frac{-\sqrt{17}-1}{4}\approx -1.280776406$$

Steps Using the Quadratic Formula

Subtract 1 from both sides.

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

Add x to both sides.

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

Combine -\frac{1}{2}x and x to get \frac{1}{2}x.

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

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

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

Square \frac{1}{2} by squaring both the numerator and the denominator of the fraction.

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

Multiply -4 times -1.

$$x=\frac{-\frac{1}{2}±\sqrt{\frac{1}{4}+4}}{2}$$

Add \frac{1}{4} to 4.

$$x=\frac{-\frac{1}{2}±\sqrt{\frac{17}{4}}}{2}$$

Take the square root of \frac{17}{4}.

$$x=\frac{-\frac{1}{2}±\frac{\sqrt{17}}{2}}{2}$$

Now solve the equation x=\frac{-\frac{1}{2}±\frac{\sqrt{17}}{2}}{2} when ± is plus. Add -\frac{1}{2} to \frac{\sqrt{17}}{2}.

$$x=\frac{\sqrt{17}-1}{2\times 2}$$

Divide \frac{-1+\sqrt{17}}{2} by 2.

$$x=\frac{\sqrt{17}-1}{4}$$

Now solve the equation x=\frac{-\frac{1}{2}±\frac{\sqrt{17}}{2}}{2} when ± is minus. Subtract \frac{\sqrt{17}}{2} from -\frac{1}{2}.

$$x=\frac{-\sqrt{17}-1}{2\times 2}$$

Divide \frac{-1-\sqrt{17}}{2} by 2.

$$x=\frac{-\sqrt{17}-1}{4}$$

The equation is now solved.

$$x=\frac{\sqrt{17}-1}{4} x=\frac{-\sqrt{17}-1}{4}$$

Steps for Completing the Square

Add x to both sides.

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

Combine -\frac{1}{2}x and x to get \frac{1}{2}x.

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

Divide \frac{1}{2}, the coefficient of the x term, by 2 to get \frac{1}{4}. Then add the square of \frac{1}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square \frac{1}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{1}{2}x+\frac{1}{16}=1+\frac{1}{16}$$

Add 1 to \frac{1}{16}.

$$x^{2}+\frac{1}{2}x+\frac{1}{16}=\frac{17}{16}$$

Factor x^{2}+\frac{1}{2}x+\frac{1}{16}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{1}{4}\right)^{2}=\frac{17}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{4}\right)^{2}}=\sqrt{\frac{17}{16}}$$

Simplify.

$$x+\frac{1}{4}=\frac{\sqrt{17}}{4} x+\frac{1}{4}=-\frac{\sqrt{17}}{4}$$

Subtract \frac{1}{4} from both sides of the equation.

$$x=\frac{\sqrt{17}-1}{4} x=\frac{-\sqrt{17}-1}{4}$$