Solve 2x-3/4x=x-1

$$\frac { 2 x - 3 } { 4 x } = x - 1$$

Solve for x (complex solution)

$$x=\frac{3+\sqrt{3}i}{4}\approx 0.75+0.433012702i$$

$$x=\frac{-\sqrt{3}i+3}{4}\approx 0.75-0.433012702i$$

Steps Using the Quadratic Formula

Variable x cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by 4x.

$$2x-3=4xx+4x\left(-1\right)$$

Multiply x and x to get x^{2}.

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

Multiply 4 and -1 to get -4.

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

Subtract 4x^{2} from both sides.

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

Add 4x to both sides.

$$2x-3-4x^{2}+4x=0$$

Combine 2x and 4x to get 6x.

$$6x-3-4x^{2}=0$$

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.

$$-4x^{2}+6x-3=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -4 for a, 6 for b, and -3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-6±\sqrt{6^{2}-4\left(-4\right)\left(-3\right)}}{2\left(-4\right)}$$

Square 6.

$$x=\frac{-6±\sqrt{36-4\left(-4\right)\left(-3\right)}}{2\left(-4\right)}$$

Multiply -4 times -4.

$$x=\frac{-6±\sqrt{36+16\left(-3\right)}}{2\left(-4\right)}$$

Multiply 16 times -3.

$$x=\frac{-6±\sqrt{36-48}}{2\left(-4\right)}$$

Add 36 to -48.

$$x=\frac{-6±\sqrt{-12}}{2\left(-4\right)}$$

Take the square root of -12.

$$x=\frac{-6±2\sqrt{3}i}{2\left(-4\right)}$$

Multiply 2 times -4.

$$x=\frac{-6±2\sqrt{3}i}{-8}$$

Now solve the equation x=\frac{-6±2\sqrt{3}i}{-8} when ± is plus. Add -6 to 2i\sqrt{3}.

$$x=\frac{-6+2\sqrt{3}i}{-8}$$

Divide -6+2i\sqrt{3} by -8.

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

Now solve the equation x=\frac{-6±2\sqrt{3}i}{-8} when ± is minus. Subtract 2i\sqrt{3} from -6.

$$x=\frac{-2\sqrt{3}i-6}{-8}$$

Divide -6-2i\sqrt{3} by -8.

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

The equation is now solved.

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

Steps for Completing the Square

Variable x cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by 4x.

$$2x-3=4xx+4x\left(-1\right)$$

Multiply x and x to get x^{2}.

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

Multiply 4 and -1 to get -4.

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

Subtract 4x^{2} from both sides.

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

Add 4x to both sides.

$$2x-3-4x^{2}+4x=0$$

Combine 2x and 4x to get 6x.

$$6x-3-4x^{2}=0$$

Add 3 to both sides. Anything plus zero gives itself.

$$6x-4x^{2}=3$$

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.

$$-4x^{2}+6x=3$$

Divide both sides by -4.

$$\frac{-4x^{2}+6x}{-4}=\frac{3}{-4}$$

Dividing by -4 undoes the multiplication by -4.

$$x^{2}+\frac{6}{-4}x=\frac{3}{-4}$$

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

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

Divide 3 by -4.

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

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.

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

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

$$x^{2}-\frac{3}{2}x+\frac{9}{16}=-\frac{3}{4}+\frac{9}{16}$$

Add -\frac{3}{4} to \frac{9}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

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

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}.

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

Take the square root of both sides of the equation.

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

Simplify.

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

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

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