Solve 440/x-450/x+10=0.5

$$\frac{ 440 }{ x } - \frac{ 450 }{ x+10 } =0.5$$

Solve for x

$$x=-110$$

$$x=80$$

Steps Using the Quadratic Formula

Variable x cannot be equal to any of the values -10,0 since division by zero is not defined. Multiply both sides of the equation by x\left(x+10\right), the least common multiple of x,x+10.

$$\left(x+10\right)\times 440-x\times 450=0.5x\left(x+10\right)$$

Use the distributive property to multiply x+10 by 440.

$$440x+4400-x\times 450=0.5x\left(x+10\right)$$

Use the distributive property to multiply 0.5x by x+10.

$$440x+4400-x\times 450=0.5x^{2}+5x$$

Subtract 0.5x^{2} from both sides.

$$440x+4400-x\times 450-0.5x^{2}=5x$$

Subtract 5x from both sides.

$$440x+4400-x\times 450-0.5x^{2}-5x=0$$

Combine 440x and -5x to get 435x.

$$435x+4400-x\times 450-0.5x^{2}=0$$

Multiply -1 and 450 to get -450.

$$435x+4400-450x-0.5x^{2}=0$$

Combine 435x and -450x to get -15x.

$$-15x+4400-0.5x^{2}=0$$

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.

$$-0.5x^{2}-15x+4400=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -0.5 for a, -15 for b, and 4400 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-15\right)±\sqrt{\left(-15\right)^{2}-4\left(-0.5\right)\times 4400}}{2\left(-0.5\right)}$$

Square -15.

$$x=\frac{-\left(-15\right)±\sqrt{225-4\left(-0.5\right)\times 4400}}{2\left(-0.5\right)}$$

Multiply -4 times -0.5.

$$x=\frac{-\left(-15\right)±\sqrt{225+2\times 4400}}{2\left(-0.5\right)}$$

Multiply 2 times 4400.

$$x=\frac{-\left(-15\right)±\sqrt{225+8800}}{2\left(-0.5\right)}$$

Add 225 to 8800.

$$x=\frac{-\left(-15\right)±\sqrt{9025}}{2\left(-0.5\right)}$$

Take the square root of 9025.

$$x=\frac{-\left(-15\right)±95}{2\left(-0.5\right)}$$

The opposite of -15 is 15.

$$x=\frac{15±95}{2\left(-0.5\right)}$$

Multiply 2 times -0.5.

$$x=\frac{15±95}{-1}$$

Now solve the equation x=\frac{15±95}{-1} when ± is plus. Add 15 to 95.

$$x=\frac{110}{-1}$$

Divide 110 by -1.

$$x=-110$$

Now solve the equation x=\frac{15±95}{-1} when ± is minus. Subtract 95 from 15.

$$x=-\frac{80}{-1}$$

Divide -80 by -1.

$$x=80$$

The equation is now solved.

$$x=-110 x=80$$

Steps for Completing the Square

Variable x cannot be equal to any of the values -10,0 since division by zero is not defined. Multiply both sides of the equation by x\left(x+10\right), the least common multiple of x,x+10.

$$\left(x+10\right)\times 440-x\times 450=0.5x\left(x+10\right)$$

Use the distributive property to multiply x+10 by 440.

$$440x+4400-x\times 450=0.5x\left(x+10\right)$$

Use the distributive property to multiply 0.5x by x+10.

$$440x+4400-x\times 450=0.5x^{2}+5x$$

Subtract 0.5x^{2} from both sides.

$$440x+4400-x\times 450-0.5x^{2}=5x$$

Subtract 5x from both sides.

$$440x+4400-x\times 450-0.5x^{2}-5x=0$$

Combine 440x and -5x to get 435x.

$$435x+4400-x\times 450-0.5x^{2}=0$$

Subtract 4400 from both sides. Anything subtracted from zero gives its negation.

$$435x-x\times 450-0.5x^{2}=-4400$$

Multiply -1 and 450 to get -450.

$$435x-450x-0.5x^{2}=-4400$$

Combine 435x and -450x to get -15x.

$$-15x-0.5x^{2}=-4400$$

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.

$$-0.5x^{2}-15x=-4400$$

Multiply both sides by -2.

$$\frac{-0.5x^{2}-15x}{-0.5}=-\frac{4400}{-0.5}$$

Dividing by -0.5 undoes the multiplication by -0.5.

$$x^{2}+\left(-\frac{15}{-0.5}\right)x=-\frac{4400}{-0.5}$$

Divide -15 by -0.5 by multiplying -15 by the reciprocal of -0.5.

$$x^{2}+30x=-\frac{4400}{-0.5}$$

Divide -4400 by -0.5 by multiplying -4400 by the reciprocal of -0.5.

$$x^{2}+30x=8800$$

Divide 30, the coefficient of the x term, by 2 to get 15. Then add the square of 15 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+30x+15^{2}=8800+15^{2}$$

Square 15.

$$x^{2}+30x+225=8800+225$$

Add 8800 to 225.

$$x^{2}+30x+225=9025$$

Factor x^{2}+30x+225. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+15\right)^{2}=9025$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+15\right)^{2}}=\sqrt{9025}$$

Simplify.

$$x+15=95 x+15=-95$$

Subtract 15 from both sides of the equation.

$$x=80 x=-110$$