Solve 10/x+3=x+9/x+4

$$\frac { 10 } { x } + 3 = \frac { x + 9 } { x + 4 }$$

Solve for x (complex solution)

$$x=\frac{-13+\sqrt{151}i}{4}\approx -3.25+3.072051432i$$

$$x=\frac{-\sqrt{151}i-13}{4}\approx -3.25-3.072051432i$$

Steps Using the Quadratic Formula

Variable x cannot be equal to any of the values -4,0 since division by zero is not defined. Multiply both sides of the equation by x\left(x+4\right), the least common multiple of x,x+4.

$$\left(x+4\right)\times 10+x\left(x+4\right)\times 3=x\left(x+9\right)$$

Use the distributive property to multiply x+4 by 10.

$$10x+40+x\left(x+4\right)\times 3=x\left(x+9\right)$$

Use the distributive property to multiply x by x+4.

$$10x+40+\left(x^{2}+4x\right)\times 3=x\left(x+9\right)$$

Use the distributive property to multiply x^{2}+4x by 3.

$$10x+40+3x^{2}+12x=x\left(x+9\right)$$

Combine 10x and 12x to get 22x.

$$22x+40+3x^{2}=x\left(x+9\right)$$

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

$$22x+40+3x^{2}=x^{2}+9x$$

Subtract x^{2} from both sides.

$$22x+40+3x^{2}-x^{2}=9x$$

Combine 3x^{2} and -x^{2} to get 2x^{2}.

$$22x+40+2x^{2}=9x$$

Subtract 9x from both sides.

$$22x+40+2x^{2}-9x=0$$

Combine 22x and -9x to get 13x.

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

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}.

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

Square 13.

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

Multiply -4 times 2.

$$x=\frac{-13±\sqrt{169-8\times 40}}{2\times 2}$$

Multiply -8 times 40.

$$x=\frac{-13±\sqrt{169-320}}{2\times 2}$$

Add 169 to -320.

$$x=\frac{-13±\sqrt{-151}}{2\times 2}$$

Take the square root of -151.

$$x=\frac{-13±\sqrt{151}i}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{-13±\sqrt{151}i}{4}$$

Now solve the equation x=\frac{-13±\sqrt{151}i}{4} when ± is plus. Add -13 to i\sqrt{151}.

$$x=\frac{-13+\sqrt{151}i}{4}$$

Now solve the equation x=\frac{-13±\sqrt{151}i}{4} when ± is minus. Subtract i\sqrt{151} from -13.

$$x=\frac{-\sqrt{151}i-13}{4}$$

The equation is now solved.

$$x=\frac{-13+\sqrt{151}i}{4} x=\frac{-\sqrt{151}i-13}{4}$$

Steps for Completing the Square

Variable x cannot be equal to any of the values -4,0 since division by zero is not defined. Multiply both sides of the equation by x\left(x+4\right), the least common multiple of x,x+4.

$$\left(x+4\right)\times 10+x\left(x+4\right)\times 3=x\left(x+9\right)$$

Use the distributive property to multiply x+4 by 10.

$$10x+40+x\left(x+4\right)\times 3=x\left(x+9\right)$$

Use the distributive property to multiply x by x+4.

$$10x+40+\left(x^{2}+4x\right)\times 3=x\left(x+9\right)$$

Use the distributive property to multiply x^{2}+4x by 3.

$$10x+40+3x^{2}+12x=x\left(x+9\right)$$

Combine 10x and 12x to get 22x.

$$22x+40+3x^{2}=x\left(x+9\right)$$

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

$$22x+40+3x^{2}=x^{2}+9x$$

Subtract x^{2} from both sides.

$$22x+40+3x^{2}-x^{2}=9x$$

Combine 3x^{2} and -x^{2} to get 2x^{2}.

$$22x+40+2x^{2}=9x$$

Subtract 9x from both sides.

$$22x+40+2x^{2}-9x=0$$

Combine 22x and -9x to get 13x.

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

Subtract 40 from both sides. Anything subtracted from zero gives its negation.

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

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.

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

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}.

$$x^{2}+\frac{13}{2}x+\frac{169}{16}=-\frac{151}{16}$$

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}.

$$\left(x+\frac{13}{4}\right)^{2}=-\frac{151}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{13}{4}\right)^{2}}=\sqrt{-\frac{151}{16}}$$

Simplify.

$$x+\frac{13}{4}=\frac{\sqrt{151}i}{4} x+\frac{13}{4}=-\frac{\sqrt{151}i}{4}$$

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

$$x=\frac{-13+\sqrt{151}i}{4} x=\frac{-\sqrt{151}i-13}{4}$$