Solve 2/x+5-x/x-5=1

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

Solve for x

$$x = \frac{\sqrt{129} - 3}{4} \approx 2.089454173$$

$$x=\frac{-\sqrt{129}-3}{4}\approx -3.589454173$$

Steps Using the Quadratic Formula

Variable x cannot be equal to any of the values -5,5 since division by zero is not defined. Multiply both sides of the equation by \left(x-5\right)\left(x+5\right), the least common multiple of x+5,x-5.

$$\left(x-5\right)\times 2-\left(x+5\right)x=\left(x-5\right)\left(x+5\right)$$

Use the distributive property to multiply x-5 by 2.

$$2x-10-\left(x+5\right)x=\left(x-5\right)\left(x+5\right)$$

Use the distributive property to multiply x+5 by x.

$$2x-10-\left(x^{2}+5x\right)=\left(x-5\right)\left(x+5\right)$$

To find the opposite of x^{2}+5x, find the opposite of each term.

$$2x-10-x^{2}-5x=\left(x-5\right)\left(x+5\right)$$

Combine 2x and -5x to get -3x.

$$-3x-10-x^{2}=\left(x-5\right)\left(x+5\right)$$

Consider \left(x-5\right)\left(x+5\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 5.

$$-3x-10-x^{2}=x^{2}-25$$

Subtract x^{2} from both sides.

$$-3x-10-x^{2}-x^{2}=-25$$

Combine -x^{2} and -x^{2} to get -2x^{2}.

$$-3x-10-2x^{2}=-25$$

Add 25 to both sides.

$$-3x-10-2x^{2}+25=0$$

Add -10 and 25 to get 15.

$$-3x+15-2x^{2}=0$$

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.

$$-2x^{2}-3x+15=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -2 for a, -3 for b, and 15 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\left(-2\right)\times 15}}{2\left(-2\right)}$$

Square -3.

$$x=\frac{-\left(-3\right)±\sqrt{9-4\left(-2\right)\times 15}}{2\left(-2\right)}$$

Multiply -4 times -2.

$$x=\frac{-\left(-3\right)±\sqrt{9+8\times 15}}{2\left(-2\right)}$$

Multiply 8 times 15.

$$x=\frac{-\left(-3\right)±\sqrt{9+120}}{2\left(-2\right)}$$

Add 9 to 120.

$$x=\frac{-\left(-3\right)±\sqrt{129}}{2\left(-2\right)}$$

The opposite of -3 is 3.

$$x=\frac{3±\sqrt{129}}{2\left(-2\right)}$$

Multiply 2 times -2.

$$x=\frac{3±\sqrt{129}}{-4}$$

Now solve the equation x=\frac{3±\sqrt{129}}{-4} when ± is plus. Add 3 to \sqrt{129}.

$$x=\frac{\sqrt{129}+3}{-4}$$

Divide 3+\sqrt{129} by -4.

$$x=\frac{-\sqrt{129}-3}{4}$$

Now solve the equation x=\frac{3±\sqrt{129}}{-4} when ± is minus. Subtract \sqrt{129} from 3.

$$x=\frac{3-\sqrt{129}}{-4}$$

Divide 3-\sqrt{129} by -4.

$$x=\frac{\sqrt{129}-3}{4}$$

The equation is now solved.

$$x=\frac{-\sqrt{129}-3}{4} x=\frac{\sqrt{129}-3}{4}$$

Steps for Completing the Square

Variable x cannot be equal to any of the values -5,5 since division by zero is not defined. Multiply both sides of the equation by \left(x-5\right)\left(x+5\right), the least common multiple of x+5,x-5.

$$\left(x-5\right)\times 2-\left(x+5\right)x=\left(x-5\right)\left(x+5\right)$$

Use the distributive property to multiply x-5 by 2.

$$2x-10-\left(x+5\right)x=\left(x-5\right)\left(x+5\right)$$

Use the distributive property to multiply x+5 by x.

$$2x-10-\left(x^{2}+5x\right)=\left(x-5\right)\left(x+5\right)$$

To find the opposite of x^{2}+5x, find the opposite of each term.

$$2x-10-x^{2}-5x=\left(x-5\right)\left(x+5\right)$$

Combine 2x and -5x to get -3x.

$$-3x-10-x^{2}=\left(x-5\right)\left(x+5\right)$$

Consider \left(x-5\right)\left(x+5\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 5.

$$-3x-10-x^{2}=x^{2}-25$$

Subtract x^{2} from both sides.

$$-3x-10-x^{2}-x^{2}=-25$$

Combine -x^{2} and -x^{2} to get -2x^{2}.

$$-3x-10-2x^{2}=-25$$

Add 10 to both sides.

$$-3x-2x^{2}=-25+10$$

Add -25 and 10 to get -15.

$$-3x-2x^{2}=-15$$

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.

$$-2x^{2}-3x=-15$$

Divide both sides by -2.

$$\frac{-2x^{2}-3x}{-2}=-\frac{15}{-2}$$

Dividing by -2 undoes the multiplication by -2.

$$x^{2}+\left(-\frac{3}{-2}\right)x=-\frac{15}{-2}$$

Divide -3 by -2.

$$x^{2}+\frac{3}{2}x=-\frac{15}{-2}$$

Divide -15 by -2.

$$x^{2}+\frac{3}{2}x=\frac{15}{2}$$

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.

$$x^{2}+\frac{3}{2}x+\left(\frac{3}{4}\right)^{2}=\frac{15}{2}+\left(\frac{3}{4}\right)^{2}$$

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

$$x^{2}+\frac{3}{2}x+\frac{9}{16}=\frac{15}{2}+\frac{9}{16}$$

Add \frac{15}{2} to \frac{9}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{3}{2}x+\frac{9}{16}=\frac{129}{16}$$

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}.

$$\left(x+\frac{3}{4}\right)^{2}=\frac{129}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{3}{4}\right)^{2}}=\sqrt{\frac{129}{16}}$$

Simplify.

$$x+\frac{3}{4}=\frac{\sqrt{129}}{4} x+\frac{3}{4}=-\frac{\sqrt{129}}{4}$$

Subtract \frac{3}{4} from both sides of the equation.

$$x=\frac{\sqrt{129}-3}{4} x=\frac{-\sqrt{129}-3}{4}$$