Solve 7=3x^2-x+2/x^2-x

$$7 = \frac { 3 x ^ { 2 } - x + 2 } { x ^ { 2 } - x }$$

Solve for x

$$x = \frac{\sqrt{17} + 3}{4} \approx 1.780776406$$

$$x=\frac{3-\sqrt{17}}{4}\approx -0.280776406$$

Steps Using the Quadratic Formula

Variable x cannot be equal to any of the values 0,1 since division by zero is not defined. Multiply both sides of the equation by x\left(x-1\right).

$$7x\left(x-1\right)=3x^{2}-x+2$$

Use the distributive property to multiply 7x by x-1.

$$7x^{2}-7x=3x^{2}-x+2$$

Subtract 3x^{2} from both sides.

$$7x^{2}-7x-3x^{2}=-x+2$$

Combine 7x^{2} and -3x^{2} to get 4x^{2}.

$$4x^{2}-7x=-x+2$$

Add x to both sides.

$$4x^{2}-7x+x=2$$

Combine -7x and x to get -6x.

$$4x^{2}-6x=2$$

Subtract 2 from both sides.

$$4x^{2}-6x-2=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, -6 for b, and -2 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-6\right)±\sqrt{\left(-6\right)^{2}-4\times 4\left(-2\right)}}{2\times 4}$$

Square -6.

$$x=\frac{-\left(-6\right)±\sqrt{36-4\times 4\left(-2\right)}}{2\times 4}$$

Multiply -4 times 4.

$$x=\frac{-\left(-6\right)±\sqrt{36-16\left(-2\right)}}{2\times 4}$$

Multiply -16 times -2.

$$x=\frac{-\left(-6\right)±\sqrt{36+32}}{2\times 4}$$

Add 36 to 32.

$$x=\frac{-\left(-6\right)±\sqrt{68}}{2\times 4}$$

Take the square root of 68.

$$x=\frac{-\left(-6\right)±2\sqrt{17}}{2\times 4}$$

The opposite of -6 is 6.

$$x=\frac{6±2\sqrt{17}}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{6±2\sqrt{17}}{8}$$

Now solve the equation x=\frac{6±2\sqrt{17}}{8} when ± is plus. Add 6 to 2\sqrt{17}.

$$x=\frac{2\sqrt{17}+6}{8}$$

Divide 6+2\sqrt{17} by 8.

$$x=\frac{\sqrt{17}+3}{4}$$

Now solve the equation x=\frac{6±2\sqrt{17}}{8} when ± is minus. Subtract 2\sqrt{17} from 6.

$$x=\frac{6-2\sqrt{17}}{8}$$

Divide 6-2\sqrt{17} by 8.

$$x=\frac{3-\sqrt{17}}{4}$$

The equation is now solved.

$$x=\frac{\sqrt{17}+3}{4} x=\frac{3-\sqrt{17}}{4}$$

Steps for Completing the Square

Variable x cannot be equal to any of the values 0,1 since division by zero is not defined. Multiply both sides of the equation by x\left(x-1\right).

$$7x\left(x-1\right)=3x^{2}-x+2$$

Use the distributive property to multiply 7x by x-1.

$$7x^{2}-7x=3x^{2}-x+2$$

Subtract 3x^{2} from both sides.

$$7x^{2}-7x-3x^{2}=-x+2$$

Combine 7x^{2} and -3x^{2} to get 4x^{2}.

$$4x^{2}-7x=-x+2$$

Add x to both sides.

$$4x^{2}-7x+x=2$$

Combine -7x and x to get -6x.

$$4x^{2}-6x=2$$

Divide both sides by 4.

$$\frac{4x^{2}-6x}{4}=\frac{2}{4}$$

Dividing by 4 undoes the multiplication by 4.

$$x^{2}+\left(-\frac{6}{4}\right)x=\frac{2}{4}$$

Reduce the fraction \frac{-6}{4} to lowest terms by extracting and canceling out 2.

$$x^{2}-\frac{3}{2}x=\frac{2}{4}$$

Reduce the fraction \frac{2}{4} to lowest terms by extracting and canceling out 2.

$$x^{2}-\frac{3}{2}x=\frac{1}{2}$$

Divide -\frac{3}{2}, the coefficient of the x term, by 2 to get -\frac{3}{4}. Then add the square of -\frac{3}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{3}{2}x+\left(-\frac{3}{4}\right)^{2}=\frac{1}{2}+\left(-\frac{3}{4}\right)^{2}$$

Square -\frac{3}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{3}{2}x+\frac{9}{16}=\frac{1}{2}+\frac{9}{16}$$

Add \frac{1}{2} to \frac{9}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{3}{2}x+\frac{9}{16}=\frac{17}{16}$$

Factor x^{2}-\frac{3}{2}x+\frac{9}{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}.

$$\left(x-\frac{3}{4}\right)^{2}=\frac{17}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{3}{4}\right)^{2}}=\sqrt{\frac{17}{16}}$$

Simplify.

$$x-\frac{3}{4}=\frac{\sqrt{17}}{4} x-\frac{3}{4}=-\frac{\sqrt{17}}{4}$$

Add \frac{3}{4} to both sides of the equation.

$$x=\frac{\sqrt{17}+3}{4} x=\frac{3-\sqrt{17}}{4}$$