Solve 3x+2/x+6-x=0

$$\frac { 3 x + 2 } { x + 6 } - x = 0$$

Solve for x

$$x=\frac{\sqrt{17}-3}{2}\approx 0.561552813$$

$$x=\frac{-\sqrt{17}-3}{2}\approx -3.561552813$$

Steps Using the Quadratic Formula

Variable x cannot be equal to -6 since division by zero is not defined. Multiply both sides of the equation by x+6.

$$3x+2-x\left(x+6\right)=0$$

Use the distributive property to multiply -x by x+6.

$$3x+2-x^{2}-6x=0$$

Combine 3x and -6x to get -3x.

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

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.

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

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, -3 for b, and 2 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\left(-1\right)\times 2}}{2\left(-1\right)}$$

Square -3.

$$x=\frac{-\left(-3\right)±\sqrt{9-4\left(-1\right)\times 2}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-\left(-3\right)±\sqrt{9+4\times 2}}{2\left(-1\right)}$$

Multiply 4 times 2.

$$x=\frac{-\left(-3\right)±\sqrt{9+8}}{2\left(-1\right)}$$

Add 9 to 8.

$$x=\frac{-\left(-3\right)±\sqrt{17}}{2\left(-1\right)}$$

The opposite of -3 is 3.

$$x=\frac{3±\sqrt{17}}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{3±\sqrt{17}}{-2}$$

Now solve the equation x=\frac{3±\sqrt{17}}{-2} when ± is plus. Add 3 to \sqrt{17}.

$$x=\frac{\sqrt{17}+3}{-2}$$

Divide 3+\sqrt{17} by -2.

$$x=\frac{-\sqrt{17}-3}{2}$$

Now solve the equation x=\frac{3±\sqrt{17}}{-2} when ± is minus. Subtract \sqrt{17} from 3.

$$x=\frac{3-\sqrt{17}}{-2}$$

Divide 3-\sqrt{17} by -2.

$$x=\frac{\sqrt{17}-3}{2}$$

The equation is now solved.

$$x=\frac{-\sqrt{17}-3}{2} x=\frac{\sqrt{17}-3}{2}$$

Steps for Completing the Square

Variable x cannot be equal to -6 since division by zero is not defined. Multiply both sides of the equation by x+6.

$$3x+2-x\left(x+6\right)=0$$

Use the distributive property to multiply -x by x+6.

$$3x+2-x^{2}-6x=0$$

Combine 3x and -6x to get -3x.

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

Subtract 2 from both sides. Anything subtracted from zero gives its negation.

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

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.

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

Divide both sides by -1.

$$\frac{-x^{2}-3x}{-1}=-\frac{2}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$x^{2}+\left(-\frac{3}{-1}\right)x=-\frac{2}{-1}$$

Divide -3 by -1.

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

Divide -2 by -1.

$$x^{2}+3x=2$$

Divide 3, the coefficient of the x term, by 2 to get \frac{3}{2}. Then add the square of \frac{3}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square \frac{3}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+3x+\frac{9}{4}=2+\frac{9}{4}$$

Add 2 to \frac{9}{4}.

$$x^{2}+3x+\frac{9}{4}=\frac{17}{4}$$

Factor x^{2}+3x+\frac{9}{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}.

$$\left(x+\frac{3}{2}\right)^{2}=\frac{17}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{3}{2}\right)^{2}}=\sqrt{\frac{17}{4}}$$

Simplify.

$$x+\frac{3}{2}=\frac{\sqrt{17}}{2} x+\frac{3}{2}=-\frac{\sqrt{17}}{2}$$

Subtract \frac{3}{2} from both sides of the equation.

$$x=\frac{\sqrt{17}-3}{2} x=\frac{-\sqrt{17}-3}{2}$$