Solve 4+x/2/x+9=x/5

$$\frac { 4 + \frac { x } { 2 } } { x + 9 } = \frac { x } { 5 }$$

Solve for x

$$x = \frac{\sqrt{489} - 13}{4} \approx 2.278336097$$

$$x=\frac{-\sqrt{489}-13}{4}\approx -8.778336097$$

Steps Using the Quadratic Formula

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

$$5\left(4+\frac{x}{2}\right)=\left(x+9\right)x$$

Use the distributive property to multiply 5 by 4+\frac{x}{2}.

$$20+5\times \frac{x}{2}=\left(x+9\right)x$$

Express 5\times \frac{x}{2} as a single fraction.

$$20+\frac{5x}{2}=\left(x+9\right)x$$

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

$$20+\frac{5x}{2}=x^{2}+9x$$

Subtract x^{2} from both sides.

$$20+\frac{5x}{2}-x^{2}=9x$$

Subtract 9x from both sides.

$$20+\frac{5x}{2}-x^{2}-9x=0$$

Multiply both sides of the equation by 2.

$$40+5x-2x^{2}-18x=0$$

Combine 5x and -18x to get -13x.

$$40-13x-2x^{2}=0$$

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.

$$-2x^{2}-13x+40=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -2 for a, -13 for b, and 40 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-13\right)±\sqrt{\left(-13\right)^{2}-4\left(-2\right)\times 40}}{2\left(-2\right)}$$

Square -13.

$$x=\frac{-\left(-13\right)±\sqrt{169-4\left(-2\right)\times 40}}{2\left(-2\right)}$$

Multiply -4 times -2.

$$x=\frac{-\left(-13\right)±\sqrt{169+8\times 40}}{2\left(-2\right)}$$

Multiply 8 times 40.

$$x=\frac{-\left(-13\right)±\sqrt{169+320}}{2\left(-2\right)}$$

Add 169 to 320.

$$x=\frac{-\left(-13\right)±\sqrt{489}}{2\left(-2\right)}$$

The opposite of -13 is 13.

$$x=\frac{13±\sqrt{489}}{2\left(-2\right)}$$

Multiply 2 times -2.

$$x=\frac{13±\sqrt{489}}{-4}$$

Now solve the equation x=\frac{13±\sqrt{489}}{-4} when ± is plus. Add 13 to \sqrt{489}.

$$x=\frac{\sqrt{489}+13}{-4}$$

Divide 13+\sqrt{489} by -4.

$$x=\frac{-\sqrt{489}-13}{4}$$

Now solve the equation x=\frac{13±\sqrt{489}}{-4} when ± is minus. Subtract \sqrt{489} from 13.

$$x=\frac{13-\sqrt{489}}{-4}$$

Divide 13-\sqrt{489} by -4.

$$x=\frac{\sqrt{489}-13}{4}$$

The equation is now solved.

$$x=\frac{-\sqrt{489}-13}{4} x=\frac{\sqrt{489}-13}{4}$$

Steps for Completing the Square

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

$$5\left(4+\frac{x}{2}\right)=\left(x+9\right)x$$

Use the distributive property to multiply 5 by 4+\frac{x}{2}.

$$20+5\times \frac{x}{2}=\left(x+9\right)x$$

Express 5\times \frac{x}{2} as a single fraction.

$$20+\frac{5x}{2}=\left(x+9\right)x$$

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

$$20+\frac{5x}{2}=x^{2}+9x$$

Subtract x^{2} from both sides.

$$20+\frac{5x}{2}-x^{2}=9x$$

Subtract 9x from both sides.

$$20+\frac{5x}{2}-x^{2}-9x=0$$

Subtract 20 from both sides. Anything subtracted from zero gives its negation.

$$\frac{5x}{2}-x^{2}-9x=-20$$

Multiply both sides of the equation by 2.

$$5x-2x^{2}-18x=-40$$

Combine 5x and -18x to get -13x.

$$-13x-2x^{2}=-40$$

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.

$$-2x^{2}-13x=-40$$

Divide both sides by -2.

$$\frac{-2x^{2}-13x}{-2}=-\frac{40}{-2}$$

Dividing by -2 undoes the multiplication by -2.

$$x^{2}+\left(-\frac{13}{-2}\right)x=-\frac{40}{-2}$$

Divide -13 by -2.

$$x^{2}+\frac{13}{2}x=-\frac{40}{-2}$$

Divide -40 by -2.

$$x^{2}+\frac{13}{2}x=20$$

Divide \frac{13}{2}, the coefficient of the x term, by 2 to get \frac{13}{4}. Then add the square of \frac{13}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{13}{2}x+\left(\frac{13}{4}\right)^{2}=20+\left(\frac{13}{4}\right)^{2}$$

Square \frac{13}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{13}{2}x+\frac{169}{16}=20+\frac{169}{16}$$

Add 20 to \frac{169}{16}.

$$x^{2}+\frac{13}{2}x+\frac{169}{16}=\frac{489}{16}$$

Factor x^{2}+\frac{13}{2}x+\frac{169}{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}.

$$\left(x+\frac{13}{4}\right)^{2}=\frac{489}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{13}{4}\right)^{2}}=\sqrt{\frac{489}{16}}$$

Simplify.

$$x+\frac{13}{4}=\frac{\sqrt{489}}{4} x+\frac{13}{4}=-\frac{\sqrt{489}}{4}$$

Subtract \frac{13}{4} from both sides of the equation.

$$x=\frac{\sqrt{489}-13}{4} x=\frac{-\sqrt{489}-13}{4}$$