Solve x+5/x^2-25=2/9

$$\frac { x + 5 } { x ^ { 2 } - 25 } = \frac { 2 } { 9 }$$

Solve for x

$$x = \frac{19}{2} = 9\frac{1}{2} = 9.5$$

Steps Using Factoring By Grouping

Variable x cannot be equal to any of the values -5,5 since division by zero is not defined. Multiply both sides of the equation by 9\left(x-5\right)\left(x+5\right), the least common multiple of x^{2}-25,9.

$$9\left(x+5\right)=2\left(x-5\right)\left(x+5\right)$$

Use the distributive property to multiply 9 by x+5.

$$9x+45=2\left(x-5\right)\left(x+5\right)$$

Use the distributive property to multiply 2 by x-5.

$$9x+45=\left(2x-10\right)\left(x+5\right)$$

Use the distributive property to multiply 2x-10 by x+5 and combine like terms.

$$9x+45=2x^{2}-50$$

Subtract 2x^{2} from both sides.

$$9x+45-2x^{2}=-50$$

Add 50 to both sides.

$$9x+45-2x^{2}+50=0$$

Add 45 and 50 to get 95.

$$9x+95-2x^{2}=0$$

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

$$-2x^{2}+9x+95=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -2x^{2}+ax+bx+95. To find a and b, set up a system to be solved.

$$a+b=9 ab=-2\times 95=-190$$

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 -190.

$$-1,190 -2,95 -5,38 -10,19$$

Calculate the sum for each pair.

$$-1+190=189 -2+95=93 -5+38=33 -10+19=9$$

The solution is the pair that gives sum 9.

$$a=19 b=-10$$

Rewrite -2x^{2}+9x+95 as \left(-2x^{2}+19x\right)+\left(-10x+95\right).

$$\left(-2x^{2}+19x\right)+\left(-10x+95\right)$$

Factor out -x in the first and -5 in the second group.

$$-x\left(2x-19\right)-5\left(2x-19\right)$$

Factor out common term 2x-19 by using distributive property.

$$\left(2x-19\right)\left(-x-5\right)$$

To find equation solutions, solve 2x-19=0 and -x-5=0.

$$x=\frac{19}{2} x=-5$$

Variable x cannot be equal to -5.

$$x=\frac{19}{2}$$

Steps Using the Quadratic Formula

Variable x cannot be equal to any of the values -5,5 since division by zero is not defined. Multiply both sides of the equation by 9\left(x-5\right)\left(x+5\right), the least common multiple of x^{2}-25,9.

$$9\left(x+5\right)=2\left(x-5\right)\left(x+5\right)$$

Use the distributive property to multiply 9 by x+5.

$$9x+45=2\left(x-5\right)\left(x+5\right)$$

Use the distributive property to multiply 2 by x-5.

$$9x+45=\left(2x-10\right)\left(x+5\right)$$

Use the distributive property to multiply 2x-10 by x+5 and combine like terms.

$$9x+45=2x^{2}-50$$

Subtract 2x^{2} from both sides.

$$9x+45-2x^{2}=-50$$

Add 50 to both sides.

$$9x+45-2x^{2}+50=0$$

Add 45 and 50 to get 95.

$$9x+95-2x^{2}=0$$

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.

$$-2x^{2}+9x+95=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -2 for a, 9 for b, and 95 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-9±\sqrt{9^{2}-4\left(-2\right)\times 95}}{2\left(-2\right)}$$

Square 9.

$$x=\frac{-9±\sqrt{81-4\left(-2\right)\times 95}}{2\left(-2\right)}$$

Multiply -4 times -2.

$$x=\frac{-9±\sqrt{81+8\times 95}}{2\left(-2\right)}$$

Multiply 8 times 95.

$$x=\frac{-9±\sqrt{81+760}}{2\left(-2\right)}$$

Add 81 to 760.

$$x=\frac{-9±\sqrt{841}}{2\left(-2\right)}$$

Take the square root of 841.

$$x=\frac{-9±29}{2\left(-2\right)}$$

Multiply 2 times -2.

$$x=\frac{-9±29}{-4}$$

Now solve the equation x=\frac{-9±29}{-4} when ± is plus. Add -9 to 29.

$$x=\frac{20}{-4}$$

Divide 20 by -4.

$$x=-5$$

Now solve the equation x=\frac{-9±29}{-4} when ± is minus. Subtract 29 from -9.

$$x=-\frac{38}{-4}$$

Reduce the fraction \frac{-38}{-4} to lowest terms by extracting and canceling out 2.

$$x=\frac{19}{2}$$

The equation is now solved.

$$x=-5 x=\frac{19}{2}$$

Variable x cannot be equal to -5.

$$x=\frac{19}{2}$$

Steps for Completing the Square

Variable x cannot be equal to any of the values -5,5 since division by zero is not defined. Multiply both sides of the equation by 9\left(x-5\right)\left(x+5\right), the least common multiple of x^{2}-25,9.

$$9\left(x+5\right)=2\left(x-5\right)\left(x+5\right)$$

Use the distributive property to multiply 9 by x+5.

$$9x+45=2\left(x-5\right)\left(x+5\right)$$

Use the distributive property to multiply 2 by x-5.

$$9x+45=\left(2x-10\right)\left(x+5\right)$$

Use the distributive property to multiply 2x-10 by x+5 and combine like terms.

$$9x+45=2x^{2}-50$$

Subtract 2x^{2} from both sides.

$$9x+45-2x^{2}=-50$$

Subtract 45 from both sides.

$$9x-2x^{2}=-50-45$$

Subtract 45 from -50 to get -95.

$$9x-2x^{2}=-95$$

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.

$$-2x^{2}+9x=-95$$

Divide both sides by -2.

$$\frac{-2x^{2}+9x}{-2}=-\frac{95}{-2}$$

Dividing by -2 undoes the multiplication by -2.

$$x^{2}+\frac{9}{-2}x=-\frac{95}{-2}$$

Divide 9 by -2.

$$x^{2}-\frac{9}{2}x=-\frac{95}{-2}$$

Divide -95 by -2.

$$x^{2}-\frac{9}{2}x=\frac{95}{2}$$

Divide -\frac{9}{2}, the coefficient of the x term, by 2 to get -\frac{9}{4}. Then add the square of -\frac{9}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{9}{2}x+\left(-\frac{9}{4}\right)^{2}=\frac{95}{2}+\left(-\frac{9}{4}\right)^{2}$$

Square -\frac{9}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{9}{2}x+\frac{81}{16}=\frac{95}{2}+\frac{81}{16}$$

Add \frac{95}{2} to \frac{81}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{9}{2}x+\frac{81}{16}=\frac{841}{16}$$

Factor x^{2}-\frac{9}{2}x+\frac{81}{16}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{9}{4}\right)^{2}=\frac{841}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{9}{4}\right)^{2}}=\sqrt{\frac{841}{16}}$$

Simplify.

$$x-\frac{9}{4}=\frac{29}{4} x-\frac{9}{4}=-\frac{29}{4}$$

Add \frac{9}{4} to both sides of the equation.

$$x=\frac{19}{2} x=-5$$

Variable x cannot be equal to -5.

$$x=\frac{19}{2}$$