Solve x-4/x+3=11/x^2+5x+6

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

Solve for x

$$x=\sqrt{10}+1\approx 4.16227766$$

$$x=1-\sqrt{10}\approx -2.16227766$$

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.

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

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

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

Multiply 1 and 1 to get 1.

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

Subtract 1 from both sides.

$$x^{2}-2x-8-1=0$$

Subtract 1 from -8 to get -9.

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

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

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

Square -2.

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

Multiply -4 times -9.

$$x=\frac{-\left(-2\right)±\sqrt{4+36}}{2}$$

Add 4 to 36.

$$x=\frac{-\left(-2\right)±\sqrt{40}}{2}$$

Take the square root of 40.

$$x=\frac{-\left(-2\right)±2\sqrt{10}}{2}$$

The opposite of -2 is 2.

$$x=\frac{2±2\sqrt{10}}{2}$$

Now solve the equation x=\frac{2±2\sqrt{10}}{2} when ± is plus. Add 2 to 2\sqrt{10}.

$$x=\frac{2\sqrt{10}+2}{2}$$

Divide 2+2\sqrt{10} by 2.

$$x=\sqrt{10}+1$$

Now solve the equation x=\frac{2±2\sqrt{10}}{2} when ± is minus. Subtract 2\sqrt{10} from 2.

$$x=\frac{2-2\sqrt{10}}{2}$$

Divide 2-2\sqrt{10} by 2.

$$x=1-\sqrt{10}$$

The equation is now solved.

$$x=\sqrt{10}+1 x=1-\sqrt{10}$$

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.

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

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

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

Multiply 1 and 1 to get 1.

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

Add 8 to both sides.

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

Add 1 and 8 to get 9.

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

Divide -2, the coefficient of the x term, by 2 to get -1. Then add the square of -1 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-2x+1=9+1$$

Add 9 to 1.

$$x^{2}-2x+1=10$$

Factor x^{2}-2x+1. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-1\right)^{2}=10$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-1\right)^{2}}=\sqrt{10}$$

Simplify.

$$x-1=\sqrt{10} x-1=-\sqrt{10}$$

Add 1 to both sides of the equation.

$$x=\sqrt{10}+1 x=1-\sqrt{10}$$