Solve 2x^2-x+1=9x^2-6x+1

$$2 x ^ { 2 } - x + 1 = 9 x ^ { 2 } - 6 x + 1$$

Solve for x

$$x=\frac{5}{7}\approx 0.714285714$$

$$x=0$$

Steps Using Factoring

Subtract 9x^{2} from both sides.

$$2x^{2}-x+1-9x^{2}=-6x+1$$

Combine 2x^{2} and -9x^{2} to get -7x^{2}.

$$-7x^{2}-x+1=-6x+1$$

Add 6x to both sides.

$$-7x^{2}-x+1+6x=1$$

Combine -x and 6x to get 5x.

$$-7x^{2}+5x+1=1$$

Subtract 1 from both sides.

$$-7x^{2}+5x+1-1=0$$

Subtract 1 from 1 to get 0.

$$-7x^{2}+5x=0$$

Factor out x.

$$x\left(-7x+5\right)=0$$

To find equation solutions, solve x=0 and -7x+5=0.

$$x=0 x=\frac{5}{7}$$

Steps Using the Quadratic Formula

Subtract 9x^{2} from both sides.

$$2x^{2}-x+1-9x^{2}=-6x+1$$

Combine 2x^{2} and -9x^{2} to get -7x^{2}.

$$-7x^{2}-x+1=-6x+1$$

Add 6x to both sides.

$$-7x^{2}-x+1+6x=1$$

Combine -x and 6x to get 5x.

$$-7x^{2}+5x+1=1$$

Subtract 1 from both sides.

$$-7x^{2}+5x+1-1=0$$

Subtract 1 from 1 to get 0.

$$-7x^{2}+5x=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -7 for a, 5 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-5±\sqrt{5^{2}}}{2\left(-7\right)}$$

Take the square root of 5^{2}.

$$x=\frac{-5±5}{2\left(-7\right)}$$

Multiply 2 times -7.

$$x=\frac{-5±5}{-14}$$

Now solve the equation x=\frac{-5±5}{-14} when ± is plus. Add -5 to 5.

$$x=\frac{0}{-14}$$

Divide 0 by -14.

$$x=0$$

Now solve the equation x=\frac{-5±5}{-14} when ± is minus. Subtract 5 from -5.

$$x=-\frac{10}{-14}$$

Reduce the fraction \frac{-10}{-14} to lowest terms by extracting and canceling out 2.

$$x=\frac{5}{7}$$

The equation is now solved.

$$x=0 x=\frac{5}{7}$$

Steps for Completing the Square

Subtract 9x^{2} from both sides.

$$2x^{2}-x+1-9x^{2}=-6x+1$$

Combine 2x^{2} and -9x^{2} to get -7x^{2}.

$$-7x^{2}-x+1=-6x+1$$

Add 6x to both sides.

$$-7x^{2}-x+1+6x=1$$

Combine -x and 6x to get 5x.

$$-7x^{2}+5x+1=1$$

Subtract 1 from both sides.

$$-7x^{2}+5x=1-1$$

Subtract 1 from 1 to get 0.

$$-7x^{2}+5x=0$$

Divide both sides by -7.

$$\frac{-7x^{2}+5x}{-7}=\frac{0}{-7}$$

Dividing by -7 undoes the multiplication by -7.

$$x^{2}+\frac{5}{-7}x=\frac{0}{-7}$$

Divide 5 by -7.

$$x^{2}-\frac{5}{7}x=\frac{0}{-7}$$

Divide 0 by -7.

$$x^{2}-\frac{5}{7}x=0$$

Divide -\frac{5}{7}, the coefficient of the x term, by 2 to get -\frac{5}{14}. Then add the square of -\frac{5}{14} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{5}{7}x+\left(-\frac{5}{14}\right)^{2}=\left(-\frac{5}{14}\right)^{2}$$

Square -\frac{5}{14} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{5}{7}x+\frac{25}{196}=\frac{25}{196}$$

Factor x^{2}-\frac{5}{7}x+\frac{25}{196}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{5}{14}\right)^{2}=\frac{25}{196}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{5}{14}\right)^{2}}=\sqrt{\frac{25}{196}}$$

Simplify.

$$x-\frac{5}{14}=\frac{5}{14} x-\frac{5}{14}=-\frac{5}{14}$$

Add \frac{5}{14} to both sides of the equation.

$$x=\frac{5}{7} x=0$$