Solve 1/x+3+1/x+8=1

$$\frac { 1 } { x + 3 } + \frac { 1 } { x + 8 } = 1$$

Solve for x

$$x=\frac{\sqrt{29}-9}{2}\approx -1.807417596$$

$$x=\frac{-\sqrt{29}-9}{2}\approx -7.192582404$$

Steps Using the Quadratic Formula

Variable x cannot be equal to any of the values -8,-3 since division by zero is not defined. Multiply both sides of the equation by \left(x+3\right)\left(x+8\right), the least common multiple of x+3,x+8.

$$x+8+x+3=\left(x+3\right)\left(x+8\right)$$

Combine x and x to get 2x.

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

Add 8 and 3 to get 11.

$$2x+11=\left(x+3\right)\left(x+8\right)$$

Use the distributive property to multiply x+3 by x+8 and combine like terms.

$$2x+11=x^{2}+11x+24$$

Subtract x^{2} from both sides.

$$2x+11-x^{2}=11x+24$$

Subtract 11x from both sides.

$$2x+11-x^{2}-11x=24$$

Combine 2x and -11x to get -9x.

$$-9x+11-x^{2}=24$$

Subtract 24 from both sides.

$$-9x+11-x^{2}-24=0$$

Subtract 24 from 11 to get -13.

$$-9x-13-x^{2}=0$$

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.

$$-x^{2}-9x-13=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, -9 for b, and -13 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-9\right)±\sqrt{\left(-9\right)^{2}-4\left(-1\right)\left(-13\right)}}{2\left(-1\right)}$$

Square -9.

$$x=\frac{-\left(-9\right)±\sqrt{81-4\left(-1\right)\left(-13\right)}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-\left(-9\right)±\sqrt{81+4\left(-13\right)}}{2\left(-1\right)}$$

Multiply 4 times -13.

$$x=\frac{-\left(-9\right)±\sqrt{81-52}}{2\left(-1\right)}$$

Add 81 to -52.

$$x=\frac{-\left(-9\right)±\sqrt{29}}{2\left(-1\right)}$$

The opposite of -9 is 9.

$$x=\frac{9±\sqrt{29}}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{9±\sqrt{29}}{-2}$$

Now solve the equation x=\frac{9±\sqrt{29}}{-2} when ± is plus. Add 9 to \sqrt{29}.

$$x=\frac{\sqrt{29}+9}{-2}$$

Divide 9+\sqrt{29} by -2.

$$x=\frac{-\sqrt{29}-9}{2}$$

Now solve the equation x=\frac{9±\sqrt{29}}{-2} when ± is minus. Subtract \sqrt{29} from 9.

$$x=\frac{9-\sqrt{29}}{-2}$$

Divide 9-\sqrt{29} by -2.

$$x=\frac{\sqrt{29}-9}{2}$$

The equation is now solved.

$$x=\frac{-\sqrt{29}-9}{2} x=\frac{\sqrt{29}-9}{2}$$

Steps for Completing the Square

Variable x cannot be equal to any of the values -8,-3 since division by zero is not defined. Multiply both sides of the equation by \left(x+3\right)\left(x+8\right), the least common multiple of x+3,x+8.

$$x+8+x+3=\left(x+3\right)\left(x+8\right)$$

Combine x and x to get 2x.

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

Add 8 and 3 to get 11.

$$2x+11=\left(x+3\right)\left(x+8\right)$$

Use the distributive property to multiply x+3 by x+8 and combine like terms.

$$2x+11=x^{2}+11x+24$$

Subtract x^{2} from both sides.

$$2x+11-x^{2}=11x+24$$

Subtract 11x from both sides.

$$2x+11-x^{2}-11x=24$$

Combine 2x and -11x to get -9x.

$$-9x+11-x^{2}=24$$

Subtract 11 from both sides.

$$-9x-x^{2}=24-11$$

Subtract 11 from 24 to get 13.

$$-9x-x^{2}=13$$

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.

$$-x^{2}-9x=13$$

Divide both sides by -1.

$$\frac{-x^{2}-9x}{-1}=\frac{13}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$x^{2}+\left(-\frac{9}{-1}\right)x=\frac{13}{-1}$$

Divide -9 by -1.

$$x^{2}+9x=\frac{13}{-1}$$

Divide 13 by -1.

$$x^{2}+9x=-13$$

Divide 9, the coefficient of the x term, by 2 to get \frac{9}{2}. Then add the square of \frac{9}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+9x+\left(\frac{9}{2}\right)^{2}=-13+\left(\frac{9}{2}\right)^{2}$$

Square \frac{9}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+9x+\frac{81}{4}=-13+\frac{81}{4}$$

Add -13 to \frac{81}{4}.

$$x^{2}+9x+\frac{81}{4}=\frac{29}{4}$$

Factor x^{2}+9x+\frac{81}{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}.

$$\left(x+\frac{9}{2}\right)^{2}=\frac{29}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{9}{2}\right)^{2}}=\sqrt{\frac{29}{4}}$$

Simplify.

$$x+\frac{9}{2}=\frac{\sqrt{29}}{2} x+\frac{9}{2}=-\frac{\sqrt{29}}{2}$$

Subtract \frac{9}{2} from both sides of the equation.

$$x=\frac{\sqrt{29}-9}{2} x=\frac{-\sqrt{29}-9}{2}$$