Solve 24x+23/x(x+2)=9/1

$$\frac { 24 x + 23 } { x ( x + 2 ) } = \frac { 9 } { 1 }$$

Solve for x

$$x = \frac{2 \sqrt{6} + 1}{3} \approx 1.966326495$$

$$x=\frac{1-2\sqrt{6}}{3}\approx -1.299659829$$

Steps Using the Quadratic Formula

Variable x cannot be equal to any of the values -2,0 since division by zero is not defined. Multiply both sides of the equation by x\left(x+2\right).

$$24x+23=x\left(x+2\right)\times 9$$

Use the distributive property to multiply x by x+2.

$$24x+23=\left(x^{2}+2x\right)\times 9$$

Use the distributive property to multiply x^{2}+2x by 9.

$$24x+23=9x^{2}+18x$$

Subtract 9x^{2} from both sides.

$$24x+23-9x^{2}=18x$$

Subtract 18x from both sides.

$$24x+23-9x^{2}-18x=0$$

Combine 24x and -18x to get 6x.

$$6x+23-9x^{2}=0$$

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.

$$-9x^{2}+6x+23=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -9 for a, 6 for b, and 23 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-6±\sqrt{6^{2}-4\left(-9\right)\times 23}}{2\left(-9\right)}$$

Square 6.

$$x=\frac{-6±\sqrt{36-4\left(-9\right)\times 23}}{2\left(-9\right)}$$

Multiply -4 times -9.

$$x=\frac{-6±\sqrt{36+36\times 23}}{2\left(-9\right)}$$

Multiply 36 times 23.

$$x=\frac{-6±\sqrt{36+828}}{2\left(-9\right)}$$

Add 36 to 828.

$$x=\frac{-6±\sqrt{864}}{2\left(-9\right)}$$

Take the square root of 864.

$$x=\frac{-6±12\sqrt{6}}{2\left(-9\right)}$$

Multiply 2 times -9.

$$x=\frac{-6±12\sqrt{6}}{-18}$$

Now solve the equation x=\frac{-6±12\sqrt{6}}{-18} when ± is plus. Add -6 to 12\sqrt{6}.

$$x=\frac{12\sqrt{6}-6}{-18}$$

Divide -6+12\sqrt{6} by -18.

$$x=\frac{1-2\sqrt{6}}{3}$$

Now solve the equation x=\frac{-6±12\sqrt{6}}{-18} when ± is minus. Subtract 12\sqrt{6} from -6.

$$x=\frac{-12\sqrt{6}-6}{-18}$$

Divide -6-12\sqrt{6} by -18.

$$x=\frac{2\sqrt{6}+1}{3}$$

The equation is now solved.

$$x=\frac{1-2\sqrt{6}}{3} x=\frac{2\sqrt{6}+1}{3}$$

Steps for Completing the Square

Variable x cannot be equal to any of the values -2,0 since division by zero is not defined. Multiply both sides of the equation by x\left(x+2\right).

$$24x+23=x\left(x+2\right)\times 9$$

Use the distributive property to multiply x by x+2.

$$24x+23=\left(x^{2}+2x\right)\times 9$$

Use the distributive property to multiply x^{2}+2x by 9.

$$24x+23=9x^{2}+18x$$

Subtract 9x^{2} from both sides.

$$24x+23-9x^{2}=18x$$

Subtract 18x from both sides.

$$24x+23-9x^{2}-18x=0$$

Combine 24x and -18x to get 6x.

$$6x+23-9x^{2}=0$$

Subtract 23 from both sides. Anything subtracted from zero gives its negation.

$$6x-9x^{2}=-23$$

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.

$$-9x^{2}+6x=-23$$

Divide both sides by -9.

$$\frac{-9x^{2}+6x}{-9}=-\frac{23}{-9}$$

Dividing by -9 undoes the multiplication by -9.

$$x^{2}+\frac{6}{-9}x=-\frac{23}{-9}$$

Reduce the fraction \frac{6}{-9} to lowest terms by extracting and canceling out 3.

$$x^{2}-\frac{2}{3}x=-\frac{23}{-9}$$

Divide -23 by -9.

$$x^{2}-\frac{2}{3}x=\frac{23}{9}$$

Divide -\frac{2}{3}, the coefficient of the x term, by 2 to get -\frac{1}{3}. Then add the square of -\frac{1}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{2}{3}x+\left(-\frac{1}{3}\right)^{2}=\frac{23}{9}+\left(-\frac{1}{3}\right)^{2}$$

Square -\frac{1}{3} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{2}{3}x+\frac{1}{9}=\frac{23+1}{9}$$

Add \frac{23}{9} to \frac{1}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{2}{3}x+\frac{1}{9}=\frac{8}{3}$$

Factor x^{2}-\frac{2}{3}x+\frac{1}{9}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{1}{3}\right)^{2}=\frac{8}{3}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{3}\right)^{2}}=\sqrt{\frac{8}{3}}$$

Simplify.

$$x-\frac{1}{3}=\frac{2\sqrt{6}}{3} x-\frac{1}{3}=-\frac{2\sqrt{6}}{3}$$

Add \frac{1}{3} to both sides of the equation.

$$x=\frac{2\sqrt{6}+1}{3} x=\frac{1-2\sqrt{6}}{3}$$