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

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

Solve for x

$$x = \frac{\sqrt{73} + 5}{2} \approx 6.772001873$$

$$x=\frac{5-\sqrt{73}}{2}\approx -1.772001873$$

Steps Using the Quadratic Formula

Variable x cannot be equal to any of the values -4,2 since division by zero is not defined. Multiply both sides of the equation by \left(x-2\right)\left(x+4\right), the least common multiple of x+4,x-2.

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

Use the distributive property to multiply x-2 by 4.

$$4x-8+\left(x+4\right)\times 3=\left(x-2\right)\left(x+4\right)$$

Use the distributive property to multiply x+4 by 3.

$$4x-8+3x+12=\left(x-2\right)\left(x+4\right)$$

Combine 4x and 3x to get 7x.

$$7x-8+12=\left(x-2\right)\left(x+4\right)$$

Add -8 and 12 to get 4.

$$7x+4=\left(x-2\right)\left(x+4\right)$$

Use the distributive property to multiply x-2 by x+4 and combine like terms.

$$7x+4=x^{2}+2x-8$$

Subtract x^{2} from both sides.

$$7x+4-x^{2}=2x-8$$

Subtract 2x from both sides.

$$7x+4-x^{2}-2x=-8$$

Combine 7x and -2x to get 5x.

$$5x+4-x^{2}=-8$$

Add 8 to both sides.

$$5x+4-x^{2}+8=0$$

Add 4 and 8 to get 12.

$$5x+12-x^{2}=0$$

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.

$$-x^{2}+5x+12=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, 5 for b, and 12 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-5±\sqrt{5^{2}-4\left(-1\right)\times 12}}{2\left(-1\right)}$$

Square 5.

$$x=\frac{-5±\sqrt{25-4\left(-1\right)\times 12}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-5±\sqrt{25+4\times 12}}{2\left(-1\right)}$$

Multiply 4 times 12.

$$x=\frac{-5±\sqrt{25+48}}{2\left(-1\right)}$$

Add 25 to 48.

$$x=\frac{-5±\sqrt{73}}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{-5±\sqrt{73}}{-2}$$

Now solve the equation x=\frac{-5±\sqrt{73}}{-2} when ± is plus. Add -5 to \sqrt{73}.

$$x=\frac{\sqrt{73}-5}{-2}$$

Divide -5+\sqrt{73} by -2.

$$x=\frac{5-\sqrt{73}}{2}$$

Now solve the equation x=\frac{-5±\sqrt{73}}{-2} when ± is minus. Subtract \sqrt{73} from -5.

$$x=\frac{-\sqrt{73}-5}{-2}$$

Divide -5-\sqrt{73} by -2.

$$x=\frac{\sqrt{73}+5}{2}$$

The equation is now solved.

$$x=\frac{5-\sqrt{73}}{2} x=\frac{\sqrt{73}+5}{2}$$

Steps for Completing the Square

Variable x cannot be equal to any of the values -4,2 since division by zero is not defined. Multiply both sides of the equation by \left(x-2\right)\left(x+4\right), the least common multiple of x+4,x-2.

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

Use the distributive property to multiply x-2 by 4.

$$4x-8+\left(x+4\right)\times 3=\left(x-2\right)\left(x+4\right)$$

Use the distributive property to multiply x+4 by 3.

$$4x-8+3x+12=\left(x-2\right)\left(x+4\right)$$

Combine 4x and 3x to get 7x.

$$7x-8+12=\left(x-2\right)\left(x+4\right)$$

Add -8 and 12 to get 4.

$$7x+4=\left(x-2\right)\left(x+4\right)$$

Use the distributive property to multiply x-2 by x+4 and combine like terms.

$$7x+4=x^{2}+2x-8$$

Subtract x^{2} from both sides.

$$7x+4-x^{2}=2x-8$$

Subtract 2x from both sides.

$$7x+4-x^{2}-2x=-8$$

Combine 7x and -2x to get 5x.

$$5x+4-x^{2}=-8$$

Subtract 4 from both sides.

$$5x-x^{2}=-8-4$$

Subtract 4 from -8 to get -12.

$$5x-x^{2}=-12$$

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.

$$-x^{2}+5x=-12$$

Divide both sides by -1.

$$\frac{-x^{2}+5x}{-1}=-\frac{12}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$x^{2}+\frac{5}{-1}x=-\frac{12}{-1}$$

Divide 5 by -1.

$$x^{2}-5x=-\frac{12}{-1}$$

Divide -12 by -1.

$$x^{2}-5x=12$$

Divide -5, the coefficient of the x term, by 2 to get -\frac{5}{2}. Then add the square of -\frac{5}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-5x+\left(-\frac{5}{2}\right)^{2}=12+\left(-\frac{5}{2}\right)^{2}$$

Square -\frac{5}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-5x+\frac{25}{4}=12+\frac{25}{4}$$

Add 12 to \frac{25}{4}.

$$x^{2}-5x+\frac{25}{4}=\frac{73}{4}$$

Factor x^{2}-5x+\frac{25}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{5}{2}\right)^{2}=\frac{73}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{5}{2}\right)^{2}}=\sqrt{\frac{73}{4}}$$

Simplify.

$$x-\frac{5}{2}=\frac{\sqrt{73}}{2} x-\frac{5}{2}=-\frac{\sqrt{73}}{2}$$

Add \frac{5}{2} to both sides of the equation.

$$x=\frac{\sqrt{73}+5}{2} x=\frac{5-\sqrt{73}}{2}$$