Solve x^2-8x^2-2x+1=0

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

Solve for x

$$x=\frac{2\sqrt{2}-1}{7}\approx 0.261203875$$

$$x=\frac{-2\sqrt{2}-1}{7}\approx -0.546918161$$

Steps Using the Quadratic Formula

Combine x^{2} and -8x^{2} to get -7x^{2}.

$$-7x^{2}-2x+1=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -7 for a, -2 for b, and 1 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-2\right)±\sqrt{\left(-2\right)^{2}-4\left(-7\right)}}{2\left(-7\right)}$$

Square -2.

$$x=\frac{-\left(-2\right)±\sqrt{4-4\left(-7\right)}}{2\left(-7\right)}$$

Multiply -4 times -7.

$$x=\frac{-\left(-2\right)±\sqrt{4+28}}{2\left(-7\right)}$$

Add 4 to 28.

$$x=\frac{-\left(-2\right)±\sqrt{32}}{2\left(-7\right)}$$

Take the square root of 32.

$$x=\frac{-\left(-2\right)±4\sqrt{2}}{2\left(-7\right)}$$

The opposite of -2 is 2.

$$x=\frac{2±4\sqrt{2}}{2\left(-7\right)}$$

Multiply 2 times -7.

$$x=\frac{2±4\sqrt{2}}{-14}$$

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

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

Divide 4\sqrt{2}+2 by -14.

$$x=\frac{-2\sqrt{2}-1}{7}$$

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

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

Divide 2-4\sqrt{2} by -14.

$$x=\frac{2\sqrt{2}-1}{7}$$

The equation is now solved.

$$x=\frac{-2\sqrt{2}-1}{7} x=\frac{2\sqrt{2}-1}{7}$$

Steps for Completing the Square

Combine x^{2} and -8x^{2} to get -7x^{2}.

$$-7x^{2}-2x+1=0$$

Subtract 1 from both sides. Anything subtracted from zero gives its negation.

$$-7x^{2}-2x=-1$$

Divide both sides by -7.

$$\frac{-7x^{2}-2x}{-7}=-\frac{1}{-7}$$

Dividing by -7 undoes the multiplication by -7.

$$x^{2}+\left(-\frac{2}{-7}\right)x=-\frac{1}{-7}$$

Divide -2 by -7.

$$x^{2}+\frac{2}{7}x=-\frac{1}{-7}$$

Divide -1 by -7.

$$x^{2}+\frac{2}{7}x=\frac{1}{7}$$

Divide \frac{2}{7}, the coefficient of the x term, by 2 to get \frac{1}{7}. Then add the square of \frac{1}{7} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{2}{7}x+\left(\frac{1}{7}\right)^{2}=\frac{1}{7}+\left(\frac{1}{7}\right)^{2}$$

Square \frac{1}{7} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{2}{7}x+\frac{1}{49}=\frac{1}{7}+\frac{1}{49}$$

Add \frac{1}{7} to \frac{1}{49} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{2}{7}x+\frac{1}{49}=\frac{8}{49}$$

Factor x^{2}+\frac{2}{7}x+\frac{1}{49}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{1}{7}\right)^{2}=\frac{8}{49}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{7}\right)^{2}}=\sqrt{\frac{8}{49}}$$

Simplify.

$$x+\frac{1}{7}=\frac{2\sqrt{2}}{7} x+\frac{1}{7}=-\frac{2\sqrt{2}}{7}$$

Subtract \frac{1}{7} from both sides of the equation.

$$x=\frac{2\sqrt{2}-1}{7} x=\frac{-2\sqrt{2}-1}{7}$$