Solve x=1+1/8*x^2-x

$$x=1+ \frac{ 1 }{ 8 } \times { x }^{ 2 } -x$$

Solve for x

$$x=2\sqrt{14}+8\approx 15.483314774$$

$$x=8-2\sqrt{14}\approx 0.516685226$$

Steps Using the Quadratic Formula

Subtract 1 from both sides.

$$x-1=\frac{1}{8}x^{2}-x$$

Subtract \frac{1}{8}x^{2} from both sides.

$$x-1-\frac{1}{8}x^{2}=-x$$

Add x to both sides.

$$x-1-\frac{1}{8}x^{2}+x=0$$

Combine x and x to get 2x.

$$2x-1-\frac{1}{8}x^{2}=0$$

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.

$$-\frac{1}{8}x^{2}+2x-1=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -\frac{1}{8} for a, 2 for b, and -1 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Square 2.

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

Multiply -4 times -\frac{1}{8}.

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

Multiply \frac{1}{2} times -1.

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

Add 4 to -\frac{1}{2}.

$$x=\frac{-2±\sqrt{\frac{7}{2}}}{2\left(-\frac{1}{8}\right)}$$

Take the square root of \frac{7}{2}.

$$x=\frac{-2±\frac{\sqrt{14}}{2}}{2\left(-\frac{1}{8}\right)}$$

Multiply 2 times -\frac{1}{8}.

$$x=\frac{-2±\frac{\sqrt{14}}{2}}{-\frac{1}{4}}$$

Now solve the equation x=\frac{-2±\frac{\sqrt{14}}{2}}{-\frac{1}{4}} when ± is plus. Add -2 to \frac{\sqrt{14}}{2}.

$$x=\frac{\frac{\sqrt{14}}{2}-2}{-\frac{1}{4}}$$

Divide -2+\frac{\sqrt{14}}{2} by -\frac{1}{4} by multiplying -2+\frac{\sqrt{14}}{2} by the reciprocal of -\frac{1}{4}.

$$x=8-2\sqrt{14}$$

Now solve the equation x=\frac{-2±\frac{\sqrt{14}}{2}}{-\frac{1}{4}} when ± is minus. Subtract \frac{\sqrt{14}}{2} from -2.

$$x=\frac{-\frac{\sqrt{14}}{2}-2}{-\frac{1}{4}}$$

Divide -2-\frac{\sqrt{14}}{2} by -\frac{1}{4} by multiplying -2-\frac{\sqrt{14}}{2} by the reciprocal of -\frac{1}{4}.

$$x=2\sqrt{14}+8$$

The equation is now solved.

$$x=8-2\sqrt{14} x=2\sqrt{14}+8$$

Steps for Completing the Square

Subtract \frac{1}{8}x^{2} from both sides.

$$x-\frac{1}{8}x^{2}=1-x$$

Add x to both sides.

$$x-\frac{1}{8}x^{2}+x=1$$

Combine x and x to get 2x.

$$2x-\frac{1}{8}x^{2}=1$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$-\frac{1}{8}x^{2}+2x=1$$

Multiply both sides by -8.

$$\frac{-\frac{1}{8}x^{2}+2x}{-\frac{1}{8}}=\frac{1}{-\frac{1}{8}}$$

Dividing by -\frac{1}{8} undoes the multiplication by -\frac{1}{8}.

$$x^{2}+\frac{2}{-\frac{1}{8}}x=\frac{1}{-\frac{1}{8}}$$

Divide 2 by -\frac{1}{8} by multiplying 2 by the reciprocal of -\frac{1}{8}.

$$x^{2}-16x=\frac{1}{-\frac{1}{8}}$$

Divide 1 by -\frac{1}{8} by multiplying 1 by the reciprocal of -\frac{1}{8}.

$$x^{2}-16x=-8$$

Divide -16, the coefficient of the x term, by 2 to get -8. Then add the square of -8 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-16x+\left(-8\right)^{2}=-8+\left(-8\right)^{2}$$

Square -8.

$$x^{2}-16x+64=-8+64$$

Add -8 to 64.

$$x^{2}-16x+64=56$$

Factor x^{2}-16x+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}.

$$\left(x-8\right)^{2}=56$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-8\right)^{2}}=\sqrt{56}$$

Simplify.

$$x-8=2\sqrt{14} x-8=-2\sqrt{14}$$

Add 8 to both sides of the equation.

$$x=2\sqrt{14}+8 x=8-2\sqrt{14}$$