Solve x^2+4x/x-3=4x+9/x-3

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

Solve for x

$$x=-3$$

Steps Using Difference of Squares

Variable x cannot be equal to 3 since division by zero is not defined. Multiply both sides of the equation by x-3.

$$x^{2}+4x=4x+9$$

Subtract 4x from both sides.

$$x^{2}+4x-4x=9$$

Combine 4x and -4x to get 0.

$$x^{2}=9$$

Subtract 9 from both sides.

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

Consider x^{2}-9. Rewrite x^{2}-9 as x^{2}-3^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

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

To find equation solutions, solve x-3=0 and x+3=0.

$$x=3 x=-3$$

Variable x cannot be equal to 3.

$$x=-3$$

Steps by Finding Square Root

Variable x cannot be equal to 3 since division by zero is not defined. Multiply both sides of the equation by x-3.

$$x^{2}+4x=4x+9$$

Subtract 4x from both sides.

$$x^{2}+4x-4x=9$$

Combine 4x and -4x to get 0.

$$x^{2}=9$$

Take the square root of both sides of the equation.

$$x=3 x=-3$$

Variable x cannot be equal to 3.

$$x=-3$$

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 x-3.

$$x^{2}+4x=4x+9$$

Subtract 4x from both sides.

$$x^{2}+4x-4x=9$$

Combine 4x and -4x to get 0.

$$x^{2}=9$$

Subtract 9 from both sides.

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

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

$$x=\frac{0±\sqrt{0^{2}-4\left(-9\right)}}{2}$$

Square 0.

$$x=\frac{0±\sqrt{-4\left(-9\right)}}{2}$$

Multiply -4 times -9.

$$x=\frac{0±\sqrt{36}}{2}$$

Take the square root of 36.

$$x=\frac{0±6}{2}$$

Now solve the equation x=\frac{0±6}{2} when ± is plus. Divide 6 by 2.

$$x=3$$

Now solve the equation x=\frac{0±6}{2} when ± is minus. Divide -6 by 2.

$$x=-3$$

The equation is now solved.

$$x=3 x=-3$$

Variable x cannot be equal to 3.

$$x=-3$$