Solve x-4/x+3=1x/x^2+5x+6

$$\frac { x - 4 } { x + 3 } = 1 \frac { x } { x ^ { 2 } + 5 x + 6 }$$

Solve for x

$$x = \frac{\sqrt{41} + 3}{2} \approx 4.701562119$$

$$x=\frac{3-\sqrt{41}}{2}\approx -1.701562119$$

Steps Using the Quadratic Formula

Variable x cannot be equal to any of the values -3,-2 since division by zero is not defined. Multiply both sides of the equation by \left(x+2\right)\left(x+3\right), the least common multiple of x+3,x^{2}+5x+6.

$$\left(x+2\right)\left(x-4\right)=1x$$

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

$$x^{2}-2x-8=1x$$

Subtract 1x from both sides.

$$x^{2}-2x-8-x=0$$

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

$$x^{2}-3x-8=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -3 for b, and -8 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\left(-8\right)}}{2}$$

Square -3.

$$x=\frac{-\left(-3\right)±\sqrt{9-4\left(-8\right)}}{2}$$

Multiply -4 times -8.

$$x=\frac{-\left(-3\right)±\sqrt{9+32}}{2}$$

Add 9 to 32.

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

The opposite of -3 is 3.

$$x=\frac{3±\sqrt{41}}{2}$$

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

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

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

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

The equation is now solved.

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

Steps for Completing the Square

Variable x cannot be equal to any of the values -3,-2 since division by zero is not defined. Multiply both sides of the equation by \left(x+2\right)\left(x+3\right), the least common multiple of x+3,x^{2}+5x+6.

$$\left(x+2\right)\left(x-4\right)=1x$$

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

$$x^{2}-2x-8=1x$$

Subtract 1x from both sides.

$$x^{2}-2x-8-x=0$$

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

$$x^{2}-3x-8=0$$

Add 8 to both sides. Anything plus zero gives itself.

$$x^{2}-3x=8$$

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.

$$x^{2}-3x+\left(-\frac{3}{2}\right)^{2}=8+\left(-\frac{3}{2}\right)^{2}$$

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

$$x^{2}-3x+\frac{9}{4}=8+\frac{9}{4}$$

Add 8 to \frac{9}{4}.

$$x^{2}-3x+\frac{9}{4}=\frac{41}{4}$$

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}.

$$\left(x-\frac{3}{2}\right)^{2}=\frac{41}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{3}{2}\right)^{2}}=\sqrt{\frac{41}{4}}$$

Simplify.

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

Add \frac{3}{2} to both sides of the equation.

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