Solve 8/x+2=17+x/12+x

$$\frac{ 8 }{ x+2 } = \frac{ 17+x }{ 12+x }$$

Solve for x

$$x = \frac{3 \sqrt{41} - 11}{2} \approx 4.104686356$$

$$x=\frac{-3\sqrt{41}-11}{2}\approx -15.104686356$$

Steps Using the Quadratic Formula

Variable x cannot be equal to any of the values -12,-2 since division by zero is not defined. Multiply both sides of the equation by \left(x+2\right)\left(x+12\right), the least common multiple of x+2,12+x.

$$\left(x+12\right)\times 8=\left(x+2\right)\left(17+x\right)$$

Use the distributive property to multiply x+12 by 8.

$$8x+96=\left(x+2\right)\left(17+x\right)$$

Use the distributive property to multiply x+2 by 17+x and combine like terms.

$$8x+96=19x+x^{2}+34$$

Subtract 19x from both sides.

$$8x+96-19x=x^{2}+34$$

Combine 8x and -19x to get -11x.

$$-11x+96=x^{2}+34$$

Subtract x^{2} from both sides.

$$-11x+96-x^{2}=34$$

Subtract 34 from both sides.

$$-11x+96-x^{2}-34=0$$

Subtract 34 from 96 to get 62.

$$-11x+62-x^{2}=0$$

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.

$$-x^{2}-11x+62=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, -11 for b, and 62 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-11\right)±\sqrt{\left(-11\right)^{2}-4\left(-1\right)\times 62}}{2\left(-1\right)}$$

Square -11.

$$x=\frac{-\left(-11\right)±\sqrt{121-4\left(-1\right)\times 62}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-\left(-11\right)±\sqrt{121+4\times 62}}{2\left(-1\right)}$$

Multiply 4 times 62.

$$x=\frac{-\left(-11\right)±\sqrt{121+248}}{2\left(-1\right)}$$

Add 121 to 248.

$$x=\frac{-\left(-11\right)±\sqrt{369}}{2\left(-1\right)}$$

Take the square root of 369.

$$x=\frac{-\left(-11\right)±3\sqrt{41}}{2\left(-1\right)}$$

The opposite of -11 is 11.

$$x=\frac{11±3\sqrt{41}}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{11±3\sqrt{41}}{-2}$$

Now solve the equation x=\frac{11±3\sqrt{41}}{-2} when ± is plus. Add 11 to 3\sqrt{41}.

$$x=\frac{3\sqrt{41}+11}{-2}$$

Divide 11+3\sqrt{41} by -2.

$$x=\frac{-3\sqrt{41}-11}{2}$$

Now solve the equation x=\frac{11±3\sqrt{41}}{-2} when ± is minus. Subtract 3\sqrt{41} from 11.

$$x=\frac{11-3\sqrt{41}}{-2}$$

Divide 11-3\sqrt{41} by -2.

$$x=\frac{3\sqrt{41}-11}{2}$$

The equation is now solved.

$$x=\frac{-3\sqrt{41}-11}{2} x=\frac{3\sqrt{41}-11}{2}$$

Steps for Completing the Square

Variable x cannot be equal to any of the values -12,-2 since division by zero is not defined. Multiply both sides of the equation by \left(x+2\right)\left(x+12\right), the least common multiple of x+2,12+x.

$$\left(x+12\right)\times 8=\left(x+2\right)\left(17+x\right)$$

Use the distributive property to multiply x+12 by 8.

$$8x+96=\left(x+2\right)\left(17+x\right)$$

Use the distributive property to multiply x+2 by 17+x and combine like terms.

$$8x+96=19x+x^{2}+34$$

Subtract 19x from both sides.

$$8x+96-19x=x^{2}+34$$

Combine 8x and -19x to get -11x.

$$-11x+96=x^{2}+34$$

Subtract x^{2} from both sides.

$$-11x+96-x^{2}=34$$

Subtract 96 from both sides.

$$-11x-x^{2}=34-96$$

Subtract 96 from 34 to get -62.

$$-11x-x^{2}=-62$$

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.

$$-x^{2}-11x=-62$$

Divide both sides by -1.

$$\frac{-x^{2}-11x}{-1}=-\frac{62}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$x^{2}+\left(-\frac{11}{-1}\right)x=-\frac{62}{-1}$$

Divide -11 by -1.

$$x^{2}+11x=-\frac{62}{-1}$$

Divide -62 by -1.

$$x^{2}+11x=62$$

Divide 11, the coefficient of the x term, by 2 to get \frac{11}{2}. Then add the square of \frac{11}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+11x+\left(\frac{11}{2}\right)^{2}=62+\left(\frac{11}{2}\right)^{2}$$

Square \frac{11}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+11x+\frac{121}{4}=62+\frac{121}{4}$$

Add 62 to \frac{121}{4}.

$$x^{2}+11x+\frac{121}{4}=\frac{369}{4}$$

Factor x^{2}+11x+\frac{121}{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}.

$$\left(x+\frac{11}{2}\right)^{2}=\frac{369}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{11}{2}\right)^{2}}=\sqrt{\frac{369}{4}}$$

Simplify.

$$x+\frac{11}{2}=\frac{3\sqrt{41}}{2} x+\frac{11}{2}=-\frac{3\sqrt{41}}{2}$$

Subtract \frac{11}{2} from both sides of the equation.

$$x=\frac{3\sqrt{41}-11}{2} x=\frac{-3\sqrt{41}-11}{2}$$