Solve 140/x+3=140/x+6

$$\frac { 140 } { x } + 3 = \frac { 140 } { x + 6 }$$

Solve for x (complex solution)

$$x=-3+\sqrt{271}i\approx -3+16.462077633i$$

$$x=-\sqrt{271}i-3\approx -3-16.462077633i$$

Steps Using the Quadratic Formula

Variable x cannot be equal to any of the values -6,0 since division by zero is not defined. Multiply both sides of the equation by x\left(x+6\right), the least common multiple of x,x+6.

$$\left(x+6\right)\times 140+x\left(x+6\right)\times 3=x\times 140$$

Use the distributive property to multiply x+6 by 140.

$$140x+840+x\left(x+6\right)\times 3=x\times 140$$

Use the distributive property to multiply x by x+6.

$$140x+840+\left(x^{2}+6x\right)\times 3=x\times 140$$

Use the distributive property to multiply x^{2}+6x by 3.

$$140x+840+3x^{2}+18x=x\times 140$$

Combine 140x and 18x to get 158x.

$$158x+840+3x^{2}=x\times 140$$

Subtract x\times 140 from both sides.

$$158x+840+3x^{2}-x\times 140=0$$

Combine 158x and -x\times 140 to get 18x.

$$18x+840+3x^{2}=0$$

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.

$$3x^{2}+18x+840=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, 18 for b, and 840 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-18±\sqrt{18^{2}-4\times 3\times 840}}{2\times 3}$$

Square 18.

$$x=\frac{-18±\sqrt{324-4\times 3\times 840}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-18±\sqrt{324-12\times 840}}{2\times 3}$$

Multiply -12 times 840.

$$x=\frac{-18±\sqrt{324-10080}}{2\times 3}$$

Add 324 to -10080.

$$x=\frac{-18±\sqrt{-9756}}{2\times 3}$$

Take the square root of -9756.

$$x=\frac{-18±6\sqrt{271}i}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{-18±6\sqrt{271}i}{6}$$

Now solve the equation x=\frac{-18±6\sqrt{271}i}{6} when ± is plus. Add -18 to 6i\sqrt{271}.

$$x=\frac{-18+6\sqrt{271}i}{6}$$

Divide -18+6i\sqrt{271} by 6.

$$x=-3+\sqrt{271}i$$

Now solve the equation x=\frac{-18±6\sqrt{271}i}{6} when ± is minus. Subtract 6i\sqrt{271} from -18.

$$x=\frac{-6\sqrt{271}i-18}{6}$$

Divide -18-6i\sqrt{271} by 6.

$$x=-\sqrt{271}i-3$$

The equation is now solved.

$$x=-3+\sqrt{271}i x=-\sqrt{271}i-3$$

Steps for Completing the Square

Variable x cannot be equal to any of the values -6,0 since division by zero is not defined. Multiply both sides of the equation by x\left(x+6\right), the least common multiple of x,x+6.

$$\left(x+6\right)\times 140+x\left(x+6\right)\times 3=x\times 140$$

Use the distributive property to multiply x+6 by 140.

$$140x+840+x\left(x+6\right)\times 3=x\times 140$$

Use the distributive property to multiply x by x+6.

$$140x+840+\left(x^{2}+6x\right)\times 3=x\times 140$$

Use the distributive property to multiply x^{2}+6x by 3.

$$140x+840+3x^{2}+18x=x\times 140$$

Combine 140x and 18x to get 158x.

$$158x+840+3x^{2}=x\times 140$$

Subtract x\times 140 from both sides.

$$158x+840+3x^{2}-x\times 140=0$$

Combine 158x and -x\times 140 to get 18x.

$$18x+840+3x^{2}=0$$

Subtract 840 from both sides. Anything subtracted from zero gives its negation.

$$18x+3x^{2}=-840$$

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.

$$3x^{2}+18x=-840$$

Divide both sides by 3.

$$\frac{3x^{2}+18x}{3}=-\frac{840}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$x^{2}+\frac{18}{3}x=-\frac{840}{3}$$

Divide 18 by 3.

$$x^{2}+6x=-\frac{840}{3}$$

Divide -840 by 3.

$$x^{2}+6x=-280$$

Divide 6, the coefficient of the x term, by 2 to get 3. Then add the square of 3 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+6x+3^{2}=-280+3^{2}$$

Square 3.

$$x^{2}+6x+9=-280+9$$

Add -280 to 9.

$$x^{2}+6x+9=-271$$

Factor x^{2}+6x+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}.

$$\left(x+3\right)^{2}=-271$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+3\right)^{2}}=\sqrt{-271}$$

Simplify.

$$x+3=\sqrt{271}i x+3=-\sqrt{271}i$$

Subtract 3 from both sides of the equation.

$$x=-3+\sqrt{271}i x=-\sqrt{271}i-3$$