Solve 3/x-4=-96/x+3/-4

$$\frac{ 3 }{ x-4 } = \frac{ - \frac{ 96 }{ x } +3 }{ -4 }$$

Solve for x

$$x=8\sqrt{2}+16\approx 27.313708499$$

$$x=16-8\sqrt{2}\approx 4.686291501$$

Steps Using the Quadratic Formula

Variable x cannot be equal to 4 since division by zero is not defined. Multiply both sides of the equation by 4\left(x-4\right), the least common multiple of x-4,-4.

$$4\times 3=\left(-x+4\right)\left(-\frac{96}{x}+3\right)$$

Multiply 4 and 3 to get 12.

$$12=\left(-x+4\right)\left(-\frac{96}{x}+3\right)$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 3 times \frac{x}{x}.

$$12=\left(-x+4\right)\left(-\frac{96}{x}+\frac{3x}{x}\right)$$

Since -\frac{96}{x} and \frac{3x}{x} have the same denominator, add them by adding their numerators.

$$12=\left(-x+4\right)\times \frac{-96+3x}{x}$$

Express \left(-x+4\right)\times \frac{-96+3x}{x} as a single fraction.

$$12=\frac{\left(-x+4\right)\left(-96+3x\right)}{x}$$

Use the distributive property to multiply -x+4 by -96+3x and combine like terms.

$$12=\frac{108x-3x^{2}-384}{x}$$

Swap sides so that all variable terms are on the left hand side.

$$\frac{108x-3x^{2}-384}{x}=12$$

Subtract 12 from both sides.

$$\frac{108x-3x^{2}-384}{x}-12=0$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 12 times \frac{x}{x}.

$$\frac{108x-3x^{2}-384}{x}-\frac{12x}{x}=0$$

Since \frac{108x-3x^{2}-384}{x} and \frac{12x}{x} have the same denominator, subtract them by subtracting their numerators.

$$\frac{108x-3x^{2}-384-12x}{x}=0$$

Combine like terms in 108x-3x^{2}-384-12x.

$$\frac{96x-3x^{2}-384}{x}=0$$

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

$$96x-3x^{2}-384=0$$

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.

$$-3x^{2}+96x-384=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -3 for a, 96 for b, and -384 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-96±\sqrt{96^{2}-4\left(-3\right)\left(-384\right)}}{2\left(-3\right)}$$

Square 96.

$$x=\frac{-96±\sqrt{9216-4\left(-3\right)\left(-384\right)}}{2\left(-3\right)}$$

Multiply -4 times -3.

$$x=\frac{-96±\sqrt{9216+12\left(-384\right)}}{2\left(-3\right)}$$

Multiply 12 times -384.

$$x=\frac{-96±\sqrt{9216-4608}}{2\left(-3\right)}$$

Add 9216 to -4608.

$$x=\frac{-96±\sqrt{4608}}{2\left(-3\right)}$$

Take the square root of 4608.

$$x=\frac{-96±48\sqrt{2}}{2\left(-3\right)}$$

Multiply 2 times -3.

$$x=\frac{-96±48\sqrt{2}}{-6}$$

Now solve the equation x=\frac{-96±48\sqrt{2}}{-6} when ± is plus. Add -96 to 48\sqrt{2}.

$$x=\frac{48\sqrt{2}-96}{-6}$$

Divide -96+48\sqrt{2} by -6.

$$x=16-8\sqrt{2}$$

Now solve the equation x=\frac{-96±48\sqrt{2}}{-6} when ± is minus. Subtract 48\sqrt{2} from -96.

$$x=\frac{-48\sqrt{2}-96}{-6}$$

Divide -96-48\sqrt{2} by -6.

$$x=8\sqrt{2}+16$$

The equation is now solved.

$$x=16-8\sqrt{2} x=8\sqrt{2}+16$$

Steps for Completing the Square

Variable x cannot be equal to 4 since division by zero is not defined. Multiply both sides of the equation by 4\left(x-4\right), the least common multiple of x-4,-4.

$$4\times 3=\left(-x+4\right)\left(-\frac{96}{x}+3\right)$$

Multiply 4 and 3 to get 12.

$$12=\left(-x+4\right)\left(-\frac{96}{x}+3\right)$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 3 times \frac{x}{x}.

$$12=\left(-x+4\right)\left(-\frac{96}{x}+\frac{3x}{x}\right)$$

Since -\frac{96}{x} and \frac{3x}{x} have the same denominator, add them by adding their numerators.

$$12=\left(-x+4\right)\times \frac{-96+3x}{x}$$

Express \left(-x+4\right)\times \frac{-96+3x}{x} as a single fraction.

$$12=\frac{\left(-x+4\right)\left(-96+3x\right)}{x}$$

Use the distributive property to multiply -x+4 by -96+3x and combine like terms.

$$12=\frac{108x-3x^{2}-384}{x}$$

Swap sides so that all variable terms are on the left hand side.

$$\frac{108x-3x^{2}-384}{x}=12$$

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

$$108x-3x^{2}-384=12x$$

Subtract 12x from both sides.

$$108x-3x^{2}-384-12x=0$$

Combine 108x and -12x to get 96x.

$$96x-3x^{2}-384=0$$

Add 384 to both sides. Anything plus zero gives itself.

$$96x-3x^{2}=384$$

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.

$$-3x^{2}+96x=384$$

Divide both sides by -3.

$$\frac{-3x^{2}+96x}{-3}=\frac{384}{-3}$$

Dividing by -3 undoes the multiplication by -3.

$$x^{2}+\frac{96}{-3}x=\frac{384}{-3}$$

Divide 96 by -3.

$$x^{2}-32x=\frac{384}{-3}$$

Divide 384 by -3.

$$x^{2}-32x=-128$$

Divide -32, the coefficient of the x term, by 2 to get -16. Then add the square of -16 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-32x+\left(-16\right)^{2}=-128+\left(-16\right)^{2}$$

Square -16.

$$x^{2}-32x+256=-128+256$$

Add -128 to 256.

$$x^{2}-32x+256=128$$

Factor x^{2}-32x+256. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-16\right)^{2}=128$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-16\right)^{2}}=\sqrt{128}$$

Simplify.

$$x-16=8\sqrt{2} x-16=-8\sqrt{2}$$

Add 16 to both sides of the equation.

$$x=8\sqrt{2}+16 x=16-8\sqrt{2}$$