Solve 648/x-3-584/x=5

$$\frac { 648 } { x - 3 } - \frac { 584 } { x } = 5$$

Solve for x

$$x = \frac{\sqrt{41281} + 79}{10} \approx 28.217726251$$

$$x=\frac{79-\sqrt{41281}}{10}\approx -12.417726251$$

Steps Using the Quadratic Formula

Variable x cannot be equal to any of the values 0,3 since division by zero is not defined. Multiply both sides of the equation by x\left(x-3\right), the least common multiple of x-3,x.

$$x\times 648-\left(x-3\right)\times 584=5x\left(x-3\right)$$

Use the distributive property to multiply x-3 by 584.

$$x\times 648-\left(584x-1752\right)=5x\left(x-3\right)$$

To find the opposite of 584x-1752, find the opposite of each term.

$$x\times 648-584x+1752=5x\left(x-3\right)$$

Combine x\times 648 and -584x to get 64x.

$$64x+1752=5x\left(x-3\right)$$

Use the distributive property to multiply 5x by x-3.

$$64x+1752=5x^{2}-15x$$

Subtract 5x^{2} from both sides.

$$64x+1752-5x^{2}=-15x$$

Add 15x to both sides.

$$64x+1752-5x^{2}+15x=0$$

Combine 64x and 15x to get 79x.

$$79x+1752-5x^{2}=0$$

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.

$$-5x^{2}+79x+1752=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -5 for a, 79 for b, and 1752 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-79±\sqrt{79^{2}-4\left(-5\right)\times 1752}}{2\left(-5\right)}$$

Square 79.

$$x=\frac{-79±\sqrt{6241-4\left(-5\right)\times 1752}}{2\left(-5\right)}$$

Multiply -4 times -5.

$$x=\frac{-79±\sqrt{6241+20\times 1752}}{2\left(-5\right)}$$

Multiply 20 times 1752.

$$x=\frac{-79±\sqrt{6241+35040}}{2\left(-5\right)}$$

Add 6241 to 35040.

$$x=\frac{-79±\sqrt{41281}}{2\left(-5\right)}$$

Multiply 2 times -5.

$$x=\frac{-79±\sqrt{41281}}{-10}$$

Now solve the equation x=\frac{-79±\sqrt{41281}}{-10} when ± is plus. Add -79 to \sqrt{41281}.

$$x=\frac{\sqrt{41281}-79}{-10}$$

Divide -79+\sqrt{41281} by -10.

$$x=\frac{79-\sqrt{41281}}{10}$$

Now solve the equation x=\frac{-79±\sqrt{41281}}{-10} when ± is minus. Subtract \sqrt{41281} from -79.

$$x=\frac{-\sqrt{41281}-79}{-10}$$

Divide -79-\sqrt{41281} by -10.

$$x=\frac{\sqrt{41281}+79}{10}$$

The equation is now solved.

$$x=\frac{79-\sqrt{41281}}{10} x=\frac{\sqrt{41281}+79}{10}$$

Steps for Completing the Square

Variable x cannot be equal to any of the values 0,3 since division by zero is not defined. Multiply both sides of the equation by x\left(x-3\right), the least common multiple of x-3,x.

$$x\times 648-\left(x-3\right)\times 584=5x\left(x-3\right)$$

Use the distributive property to multiply x-3 by 584.

$$x\times 648-\left(584x-1752\right)=5x\left(x-3\right)$$

To find the opposite of 584x-1752, find the opposite of each term.

$$x\times 648-584x+1752=5x\left(x-3\right)$$

Combine x\times 648 and -584x to get 64x.

$$64x+1752=5x\left(x-3\right)$$

Use the distributive property to multiply 5x by x-3.

$$64x+1752=5x^{2}-15x$$

Subtract 5x^{2} from both sides.

$$64x+1752-5x^{2}=-15x$$

Add 15x to both sides.

$$64x+1752-5x^{2}+15x=0$$

Combine 64x and 15x to get 79x.

$$79x+1752-5x^{2}=0$$

Subtract 1752 from both sides. Anything subtracted from zero gives its negation.

$$79x-5x^{2}=-1752$$

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.

$$-5x^{2}+79x=-1752$$

Divide both sides by -5.

$$\frac{-5x^{2}+79x}{-5}=-\frac{1752}{-5}$$

Dividing by -5 undoes the multiplication by -5.

$$x^{2}+\frac{79}{-5}x=-\frac{1752}{-5}$$

Divide 79 by -5.

$$x^{2}-\frac{79}{5}x=-\frac{1752}{-5}$$

Divide -1752 by -5.

$$x^{2}-\frac{79}{5}x=\frac{1752}{5}$$

Divide -\frac{79}{5}, the coefficient of the x term, by 2 to get -\frac{79}{10}. Then add the square of -\frac{79}{10} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{79}{5}x+\left(-\frac{79}{10}\right)^{2}=\frac{1752}{5}+\left(-\frac{79}{10}\right)^{2}$$

Square -\frac{79}{10} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{79}{5}x+\frac{6241}{100}=\frac{1752}{5}+\frac{6241}{100}$$

Add \frac{1752}{5} to \frac{6241}{100} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{79}{5}x+\frac{6241}{100}=\frac{41281}{100}$$

Factor x^{2}-\frac{79}{5}x+\frac{6241}{100}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{79}{10}\right)^{2}=\frac{41281}{100}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{79}{10}\right)^{2}}=\sqrt{\frac{41281}{100}}$$

Simplify.

$$x-\frac{79}{10}=\frac{\sqrt{41281}}{10} x-\frac{79}{10}=-\frac{\sqrt{41281}}{10}$$

Add \frac{79}{10} to both sides of the equation.

$$x=\frac{\sqrt{41281}+79}{10} x=\frac{79-\sqrt{41281}}{10}$$