Solve 1/x-3-x+4/x^2-9=1

$$\frac { 1 } { x - 3 } - \frac { x + 4 } { x ^ { 2 } - 9 } = 1$$

Solve for x

$$x=2\sqrt{2}\approx 2.828427125$$

$$x=-2\sqrt{2}\approx -2.828427125$$

Steps by Finding Square Root

Variable x cannot be equal to any of the values -3,3 since division by zero is not defined. Multiply both sides of the equation by \left(x-3\right)\left(x+3\right), the least common multiple of x-3,x^{2}-9.

$$x+3-\left(x+4\right)=\left(x-3\right)\left(x+3\right)$$

To find the opposite of x+4, find the opposite of each term.

$$x+3-x-4=\left(x-3\right)\left(x+3\right)$$

Combine x and -x to get 0.

$$3-4=\left(x-3\right)\left(x+3\right)$$

Subtract 4 from 3 to get -1.

$$-1=\left(x-3\right)\left(x+3\right)$$

Consider \left(x-3\right)\left(x+3\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 3.

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

Swap sides so that all variable terms are on the left hand side.

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

Add 9 to both sides.

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

Add -1 and 9 to get 8.

$$x^{2}=8$$

Take the square root of both sides of the equation.

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

Steps Using the Quadratic Formula

Variable x cannot be equal to any of the values -3,3 since division by zero is not defined. Multiply both sides of the equation by \left(x-3\right)\left(x+3\right), the least common multiple of x-3,x^{2}-9.

$$x+3-\left(x+4\right)=\left(x-3\right)\left(x+3\right)$$

To find the opposite of x+4, find the opposite of each term.

$$x+3-x-4=\left(x-3\right)\left(x+3\right)$$

Combine x and -x to get 0.

$$3-4=\left(x-3\right)\left(x+3\right)$$

Subtract 4 from 3 to get -1.

$$-1=\left(x-3\right)\left(x+3\right)$$

Consider \left(x-3\right)\left(x+3\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 3.

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

Swap sides so that all variable terms are on the left hand side.

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

Add 1 to both sides.

$$x^{2}-9+1=0$$

Add -9 and 1 to get -8.

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

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

$$x=\frac{0±\sqrt{0^{2}-4\left(-8\right)}}{2}$$

Square 0.

$$x=\frac{0±\sqrt{-4\left(-8\right)}}{2}$$

Multiply -4 times -8.

$$x=\frac{0±\sqrt{32}}{2}$$

Take the square root of 32.

$$x=\frac{0±4\sqrt{2}}{2}$$

Now solve the equation x=\frac{0±4\sqrt{2}}{2} when ± is plus.

$$x=2\sqrt{2}$$

Now solve the equation x=\frac{0±4\sqrt{2}}{2} when ± is minus.

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

The equation is now solved.

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