Solve 4x-1/x=3x

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

Solve for x

$$x=1$$

$$x=\frac{1}{3}\approx 0.333333333$$

Steps Using Factoring By Grouping

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

$$4x-1=3xx$$

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

$$4x-1=3x^{2}$$

Subtract 3x^{2} from both sides.

$$4x-1-3x^{2}=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$-3x^{2}+4x-1=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -3x^{2}+ax+bx-1. To find a and b, set up a system to be solved.

$$a+b=4 ab=-3\left(-1\right)=3$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. The only such pair is the system solution.

$$a=3 b=1$$

Rewrite -3x^{2}+4x-1 as \left(-3x^{2}+3x\right)+\left(x-1\right).

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

Factor out 3x in the first and -1 in the second group.

$$3x\left(-x+1\right)-\left(-x+1\right)$$

Factor out common term -x+1 by using distributive property.

$$\left(-x+1\right)\left(3x-1\right)$$

To find equation solutions, solve -x+1=0 and 3x-1=0.

$$x=1 x=\frac{1}{3}$$

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 x.

$$4x-1=3xx$$

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

$$4x-1=3x^{2}$$

Subtract 3x^{2} from both sides.

$$4x-1-3x^{2}=0$$

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.

$$-3x^{2}+4x-1=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -3 for a, 4 for b, and -1 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Square 4.

$$x=\frac{-4±\sqrt{16-4\left(-3\right)\left(-1\right)}}{2\left(-3\right)}$$

Multiply -4 times -3.

$$x=\frac{-4±\sqrt{16+12\left(-1\right)}}{2\left(-3\right)}$$

Multiply 12 times -1.

$$x=\frac{-4±\sqrt{16-12}}{2\left(-3\right)}$$

Add 16 to -12.

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

Take the square root of 4.

$$x=\frac{-4±2}{2\left(-3\right)}$$

Multiply 2 times -3.

$$x=\frac{-4±2}{-6}$$

Now solve the equation x=\frac{-4±2}{-6} when ± is plus. Add -4 to 2.

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

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

$$x=\frac{1}{3}$$

Now solve the equation x=\frac{-4±2}{-6} when ± is minus. Subtract 2 from -4.

$$x=-\frac{6}{-6}$$

Divide -6 by -6.

$$x=1$$

The equation is now solved.

$$x=\frac{1}{3} x=1$$

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 x.

$$4x-1=3xx$$

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

$$4x-1=3x^{2}$$

Subtract 3x^{2} from both sides.

$$4x-1-3x^{2}=0$$

Add 1 to both sides. Anything plus zero gives itself.

$$4x-3x^{2}=1$$

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.

$$-3x^{2}+4x=1$$

Divide both sides by -3.

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

Dividing by -3 undoes the multiplication by -3.

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

Divide 4 by -3.

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

Divide 1 by -3.

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

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

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

Square -\frac{2}{3} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{4}{3}x+\frac{4}{9}=-\frac{1}{3}+\frac{4}{9}$$

Add -\frac{1}{3} to \frac{4}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{4}{3}x+\frac{4}{9}=\frac{1}{9}$$

Factor x^{2}-\frac{4}{3}x+\frac{4}{9}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{2}{3}\right)^{2}=\frac{1}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{2}{3}\right)^{2}}=\sqrt{\frac{1}{9}}$$

Simplify.

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

Add \frac{2}{3} to both sides of the equation.

$$x=1 x=\frac{1}{3}$$