Solve x+3/4=9/x+3

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

Solve for x

$$x=-9$$

$$x=3$$

Steps Using the Quadratic Formula

Variable x cannot be equal to -3 since division by zero is not defined. Multiply both sides of the equation by 4\left(x+3\right), the least common multiple of 4,x+3.

$$\left(x+3\right)\left(x+3\right)=4\times 9$$

Multiply x+3 and x+3 to get \left(x+3\right)^{2}.

$$\left(x+3\right)^{2}=4\times 9$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+3\right)^{2}.

$$x^{2}+6x+9=4\times 9$$

Multiply 4 and 9 to get 36.

$$x^{2}+6x+9=36$$

Subtract 36 from both sides.

$$x^{2}+6x+9-36=0$$

Subtract 36 from 9 to get -27.

$$x^{2}+6x-27=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 6 for b, and -27 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-6±\sqrt{6^{2}-4\left(-27\right)}}{2}$$

Square 6.

$$x=\frac{-6±\sqrt{36-4\left(-27\right)}}{2}$$

Multiply -4 times -27.

$$x=\frac{-6±\sqrt{36+108}}{2}$$

Add 36 to 108.

$$x=\frac{-6±\sqrt{144}}{2}$$

Take the square root of 144.

$$x=\frac{-6±12}{2}$$

Now solve the equation x=\frac{-6±12}{2} when ± is plus. Add -6 to 12.

$$x=\frac{6}{2}$$

Divide 6 by 2.

$$x=3$$

Now solve the equation x=\frac{-6±12}{2} when ± is minus. Subtract 12 from -6.

$$x=-\frac{18}{2}$$

Divide -18 by 2.

$$x=-9$$

The equation is now solved.

$$x=3 x=-9$$

Steps for Completing the Square

Variable x cannot be equal to -3 since division by zero is not defined. Multiply both sides of the equation by 4\left(x+3\right), the least common multiple of 4,x+3.

$$\left(x+3\right)\left(x+3\right)=4\times 9$$

Multiply x+3 and x+3 to get \left(x+3\right)^{2}.

$$\left(x+3\right)^{2}=4\times 9$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+3\right)^{2}.

$$x^{2}+6x+9=4\times 9$$

Multiply 4 and 9 to get 36.

$$x^{2}+6x+9=36$$

Factor x^{2}+6x+9. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+3\right)^{2}=36$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+3\right)^{2}}=\sqrt{36}$$

Simplify.

$$x+3=6 x+3=-6$$

Subtract 3 from both sides of the equation.

$$x=3 x=-9$$