Solve 3x-4/x=4-7x

$$\frac { 3 x - 4 } { x } = 4 - 7 x$$

Solve for x

$$x=\frac{\sqrt{113}+1}{14}\approx 0.830724701$$

$$x=\frac{1-\sqrt{113}}{14}\approx -0.687867558$$

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.

$$3x-4=x\times 4-7xx$$

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

$$3x-4=x\times 4-7x^{2}$$

Subtract x\times 4 from both sides.

$$3x-4-x\times 4=-7x^{2}$$

Combine 3x and -x\times 4 to get -x.

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

Add 7x^{2} to both sides.

$$-x-4+7x^{2}=0$$

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.

$$7x^{2}-x-4=0$$

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

$$x=\frac{-\left(-1\right)±\sqrt{1-4\times 7\left(-4\right)}}{2\times 7}$$

Multiply -4 times 7.

$$x=\frac{-\left(-1\right)±\sqrt{1-28\left(-4\right)}}{2\times 7}$$

Multiply -28 times -4.

$$x=\frac{-\left(-1\right)±\sqrt{1+112}}{2\times 7}$$

Add 1 to 112.

$$x=\frac{-\left(-1\right)±\sqrt{113}}{2\times 7}$$

The opposite of -1 is 1.

$$x=\frac{1±\sqrt{113}}{2\times 7}$$

Multiply 2 times 7.

$$x=\frac{1±\sqrt{113}}{14}$$

Now solve the equation x=\frac{1±\sqrt{113}}{14} when ± is plus. Add 1 to \sqrt{113}.

$$x=\frac{\sqrt{113}+1}{14}$$

Now solve the equation x=\frac{1±\sqrt{113}}{14} when ± is minus. Subtract \sqrt{113} from 1.

$$x=\frac{1-\sqrt{113}}{14}$$

The equation is now solved.

$$x=\frac{\sqrt{113}+1}{14} x=\frac{1-\sqrt{113}}{14}$$

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.

$$3x-4=x\times 4-7xx$$

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

$$3x-4=x\times 4-7x^{2}$$

Subtract x\times 4 from both sides.

$$3x-4-x\times 4=-7x^{2}$$

Combine 3x and -x\times 4 to get -x.

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

Add 7x^{2} to both sides.

$$-x-4+7x^{2}=0$$

Add 4 to both sides. Anything plus zero gives itself.

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

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.

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

Divide both sides by 7.

$$\frac{7x^{2}-x}{7}=\frac{4}{7}$$

Dividing by 7 undoes the multiplication by 7.

$$x^{2}-\frac{1}{7}x=\frac{4}{7}$$

Divide -\frac{1}{7}, the coefficient of the x term, by 2 to get -\frac{1}{14}. Then add the square of -\frac{1}{14} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{1}{7}x+\left(-\frac{1}{14}\right)^{2}=\frac{4}{7}+\left(-\frac{1}{14}\right)^{2}$$

Square -\frac{1}{14} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{1}{7}x+\frac{1}{196}=\frac{4}{7}+\frac{1}{196}$$

Add \frac{4}{7} to \frac{1}{196} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{1}{7}x+\frac{1}{196}=\frac{113}{196}$$

Factor x^{2}-\frac{1}{7}x+\frac{1}{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}.

$$\left(x-\frac{1}{14}\right)^{2}=\frac{113}{196}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{14}\right)^{2}}=\sqrt{\frac{113}{196}}$$

Simplify.

$$x-\frac{1}{14}=\frac{\sqrt{113}}{14} x-\frac{1}{14}=-\frac{\sqrt{113}}{14}$$

Add \frac{1}{14} to both sides of the equation.

$$x=\frac{\sqrt{113}+1}{14} x=\frac{1-\sqrt{113}}{14}$$