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

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

Solve for x

$$x=2\sqrt{2}+1\approx 3.828427125$$

$$x=1-2\sqrt{2}\approx -1.828427125$$

Steps Using the Quadratic Formula

Variable x cannot be equal to any of the values -2,2 since division by zero is not defined. Multiply both sides of the equation by \left(x-2\right)\left(x+2\right).

$$2x+3=\left(x-2\right)\left(x+2\right)$$

Consider \left(x-2\right)\left(x+2\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 2.

$$2x+3=x^{2}-4$$

Subtract x^{2} from both sides.

$$2x+3-x^{2}=-4$$

Add 4 to both sides.

$$2x+3-x^{2}+4=0$$

Add 3 and 4 to get 7.

$$2x+7-x^{2}=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$-x^{2}+2x+7=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, 2 for b, and 7 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-2±\sqrt{2^{2}-4\left(-1\right)\times 7}}{2\left(-1\right)}$$

Square 2.

$$x=\frac{-2±\sqrt{4-4\left(-1\right)\times 7}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-2±\sqrt{4+4\times 7}}{2\left(-1\right)}$$

Multiply 4 times 7.

$$x=\frac{-2±\sqrt{4+28}}{2\left(-1\right)}$$

Add 4 to 28.

$$x=\frac{-2±\sqrt{32}}{2\left(-1\right)}$$

Take the square root of 32.

$$x=\frac{-2±4\sqrt{2}}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{-2±4\sqrt{2}}{-2}$$

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

$$x=\frac{4\sqrt{2}-2}{-2}$$

Divide 4\sqrt{2}-2 by -2.

$$x=1-2\sqrt{2}$$

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

$$x=\frac{-4\sqrt{2}-2}{-2}$$

Divide -2-4\sqrt{2} by -2.

$$x=2\sqrt{2}+1$$

The equation is now solved.

$$x=1-2\sqrt{2} x=2\sqrt{2}+1$$

Steps for Completing the Square

Variable x cannot be equal to any of the values -2,2 since division by zero is not defined. Multiply both sides of the equation by \left(x-2\right)\left(x+2\right).

$$2x+3=\left(x-2\right)\left(x+2\right)$$

Consider \left(x-2\right)\left(x+2\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 2.

$$2x+3=x^{2}-4$$

Subtract x^{2} from both sides.

$$2x+3-x^{2}=-4$$

Subtract 3 from both sides.

$$2x-x^{2}=-4-3$$

Subtract 3 from -4 to get -7.

$$2x-x^{2}=-7$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$-x^{2}+2x=-7$$

Divide both sides by -1.

$$\frac{-x^{2}+2x}{-1}=-\frac{7}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$x^{2}+\frac{2}{-1}x=-\frac{7}{-1}$$

Divide 2 by -1.

$$x^{2}-2x=-\frac{7}{-1}$$

Divide -7 by -1.

$$x^{2}-2x=7$$

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.

$$x^{2}-2x+1=7+1$$

Add 7 to 1.

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

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}.

$$\left(x-1\right)^{2}=8$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-1\right)^{2}}=\sqrt{8}$$

Simplify.

$$x-1=2\sqrt{2} x-1=-2\sqrt{2}$$

Add 1 to both sides of the equation.

$$x=2\sqrt{2}+1 x=1-2\sqrt{2}$$