Solve 5x+14/x^2-4=x^2/x^2-4

$$\frac{ 5x+14 }{ { x }^{ 2 } -4 } = \frac{ { x }^{ 2 } }{ { x }^{ 2 } -4 }$$

Solve for x

$$x=7$$

Steps Using Factoring By Grouping

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).

$$5x+14=x^{2}$$

Subtract x^{2} from both sides.

$$5x+14-x^{2}=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$-x^{2}+5x+14=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx+14. To find a and b, set up a system to be solved.

$$a+b=5 ab=-14=-14$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -14.

$$-1,14 -2,7$$

Calculate the sum for each pair.

$$-1+14=13 -2+7=5$$

The solution is the pair that gives sum 5.

$$a=7 b=-2$$

Rewrite -x^{2}+5x+14 as \left(-x^{2}+7x\right)+\left(-2x+14\right).

$$\left(-x^{2}+7x\right)+\left(-2x+14\right)$$

Factor out -x in the first and -2 in the second group.

$$-x\left(x-7\right)-2\left(x-7\right)$$

Factor out common term x-7 by using distributive property.

$$\left(x-7\right)\left(-x-2\right)$$

To find equation solutions, solve x-7=0 and -x-2=0.

$$x=7 x=-2$$

Variable x cannot be equal to -2.

$$x=7$$

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).

$$5x+14=x^{2}$$

Subtract x^{2} from both sides.

$$5x+14-x^{2}=0$$

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.

$$-x^{2}+5x+14=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, 5 for b, and 14 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-5±\sqrt{5^{2}-4\left(-1\right)\times 14}}{2\left(-1\right)}$$

Square 5.

$$x=\frac{-5±\sqrt{25-4\left(-1\right)\times 14}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-5±\sqrt{25+4\times 14}}{2\left(-1\right)}$$

Multiply 4 times 14.

$$x=\frac{-5±\sqrt{25+56}}{2\left(-1\right)}$$

Add 25 to 56.

$$x=\frac{-5±\sqrt{81}}{2\left(-1\right)}$$

Take the square root of 81.

$$x=\frac{-5±9}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{-5±9}{-2}$$

Now solve the equation x=\frac{-5±9}{-2} when ± is plus. Add -5 to 9.

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

Divide 4 by -2.

$$x=-2$$

Now solve the equation x=\frac{-5±9}{-2} when ± is minus. Subtract 9 from -5.

$$x=-\frac{14}{-2}$$

Divide -14 by -2.

$$x=7$$

The equation is now solved.

$$x=-2 x=7$$

Variable x cannot be equal to -2.

$$x=7$$

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).

$$5x+14=x^{2}$$

Subtract x^{2} from both sides.

$$5x+14-x^{2}=0$$

Subtract 14 from both sides. Anything subtracted from zero gives its negation.

$$5x-x^{2}=-14$$

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.

$$-x^{2}+5x=-14$$

Divide both sides by -1.

$$\frac{-x^{2}+5x}{-1}=-\frac{14}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$x^{2}+\frac{5}{-1}x=-\frac{14}{-1}$$

Divide 5 by -1.

$$x^{2}-5x=-\frac{14}{-1}$$

Divide -14 by -1.

$$x^{2}-5x=14$$

Divide -5, the coefficient of the x term, by 2 to get -\frac{5}{2}. Then add the square of -\frac{5}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-5x+\left(-\frac{5}{2}\right)^{2}=14+\left(-\frac{5}{2}\right)^{2}$$

Square -\frac{5}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-5x+\frac{25}{4}=14+\frac{25}{4}$$

Add 14 to \frac{25}{4}.

$$x^{2}-5x+\frac{25}{4}=\frac{81}{4}$$

Factor x^{2}-5x+\frac{25}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{5}{2}\right)^{2}=\frac{81}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{5}{2}\right)^{2}}=\sqrt{\frac{81}{4}}$$

Simplify.

$$x-\frac{5}{2}=\frac{9}{2} x-\frac{5}{2}=-\frac{9}{2}$$

Add \frac{5}{2} to both sides of the equation.

$$x=7 x=-2$$

Variable x cannot be equal to -2.

$$x=7$$