Solve 5/x-3/2=x/5

$$\frac { 5 } { x } - \frac { 3 } { 2 } = \frac { x } { 5 }$$

Solve for x

$$x = \frac{5}{2} = 2\frac{1}{2} = 2.5$$

$$x=-10$$

Steps Using Factoring By Grouping

Variable x cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by 10x, the least common multiple of x,2,5.

$$10\times 5+10x\left(-\frac{3}{2}\right)=2xx$$

Multiply 10 and 5 to get 50.

$$50+10x\left(-\frac{3}{2}\right)=2xx$$

Express 10\left(-\frac{3}{2}\right) as a single fraction.

$$50+\frac{10\left(-3\right)}{2}x=2xx$$

Multiply 10 and -3 to get -30.

$$50+\frac{-30}{2}x=2xx$$

Divide -30 by 2 to get -15.

$$50-15x=2xx$$

Multiply x and x to get x^{2}.

$$50-15x=2x^{2}$$

Subtract 2x^{2} from both sides.

$$50-15x-2x^{2}=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$-2x^{2}-15x+50=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -2x^{2}+ax+bx+50. To find a and b, set up a system to be solved.

$$a+b=-15 ab=-2\times 50=-100$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -100.

$$1,-100 2,-50 4,-25 5,-20 10,-10$$

Calculate the sum for each pair.

$$1-100=-99 2-50=-48 4-25=-21 5-20=-15 10-10=0$$

The solution is the pair that gives sum -15.

$$a=5 b=-20$$

Rewrite -2x^{2}-15x+50 as \left(-2x^{2}+5x\right)+\left(-20x+50\right).

$$\left(-2x^{2}+5x\right)+\left(-20x+50\right)$$

Factor out -x in the first and -10 in the second group.

$$-x\left(2x-5\right)-10\left(2x-5\right)$$

Factor out common term 2x-5 by using distributive property.

$$\left(2x-5\right)\left(-x-10\right)$$

To find equation solutions, solve 2x-5=0 and -x-10=0.

$$x=\frac{5}{2} x=-10$$

Steps Using the Quadratic Formula

Variable x cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by 10x, the least common multiple of x,2,5.

$$10\times 5+10x\left(-\frac{3}{2}\right)=2xx$$

Multiply 10 and 5 to get 50.

$$50+10x\left(-\frac{3}{2}\right)=2xx$$

Express 10\left(-\frac{3}{2}\right) as a single fraction.

$$50+\frac{10\left(-3\right)}{2}x=2xx$$

Multiply 10 and -3 to get -30.

$$50+\frac{-30}{2}x=2xx$$

Divide -30 by 2 to get -15.

$$50-15x=2xx$$

Multiply x and x to get x^{2}.

$$50-15x=2x^{2}$$

Subtract 2x^{2} from both sides.

$$50-15x-2x^{2}=0$$

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.

$$-2x^{2}-15x+50=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -2 for a, -15 for b, and 50 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-15\right)±\sqrt{\left(-15\right)^{2}-4\left(-2\right)\times 50}}{2\left(-2\right)}$$

Square -15.

$$x=\frac{-\left(-15\right)±\sqrt{225-4\left(-2\right)\times 50}}{2\left(-2\right)}$$

Multiply -4 times -2.

$$x=\frac{-\left(-15\right)±\sqrt{225+8\times 50}}{2\left(-2\right)}$$

Multiply 8 times 50.

$$x=\frac{-\left(-15\right)±\sqrt{225+400}}{2\left(-2\right)}$$

Add 225 to 400.

$$x=\frac{-\left(-15\right)±\sqrt{625}}{2\left(-2\right)}$$

Take the square root of 625.

$$x=\frac{-\left(-15\right)±25}{2\left(-2\right)}$$

The opposite of -15 is 15.

$$x=\frac{15±25}{2\left(-2\right)}$$

Multiply 2 times -2.

$$x=\frac{15±25}{-4}$$

Now solve the equation x=\frac{15±25}{-4} when ± is plus. Add 15 to 25.

$$x=\frac{40}{-4}$$

Divide 40 by -4.

$$x=-10$$

Now solve the equation x=\frac{15±25}{-4} when ± is minus. Subtract 25 from 15.

$$x=-\frac{10}{-4}$$

Reduce the fraction \frac{-10}{-4} to lowest terms by extracting and canceling out 2.

$$x=\frac{5}{2}$$

The equation is now solved.

$$x=-10 x=\frac{5}{2}$$

Steps for Completing the Square

Variable x cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by 10x, the least common multiple of x,2,5.

$$10\times 5+10x\left(-\frac{3}{2}\right)=2xx$$

Multiply 10 and 5 to get 50.

$$50+10x\left(-\frac{3}{2}\right)=2xx$$

Express 10\left(-\frac{3}{2}\right) as a single fraction.

$$50+\frac{10\left(-3\right)}{2}x=2xx$$

Multiply 10 and -3 to get -30.

$$50+\frac{-30}{2}x=2xx$$

Divide -30 by 2 to get -15.

$$50-15x=2xx$$

Multiply x and x to get x^{2}.

$$50-15x=2x^{2}$$

Subtract 2x^{2} from both sides.

$$50-15x-2x^{2}=0$$

Subtract 50 from both sides. Anything subtracted from zero gives its negation.

$$-15x-2x^{2}=-50$$

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.

$$-2x^{2}-15x=-50$$

Divide both sides by -2.

$$\frac{-2x^{2}-15x}{-2}=-\frac{50}{-2}$$

Dividing by -2 undoes the multiplication by -2.

$$x^{2}+\left(-\frac{15}{-2}\right)x=-\frac{50}{-2}$$

Divide -15 by -2.

$$x^{2}+\frac{15}{2}x=-\frac{50}{-2}$$

Divide -50 by -2.

$$x^{2}+\frac{15}{2}x=25$$

Divide \frac{15}{2}, the coefficient of the x term, by 2 to get \frac{15}{4}. Then add the square of \frac{15}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{15}{2}x+\left(\frac{15}{4}\right)^{2}=25+\left(\frac{15}{4}\right)^{2}$$

Square \frac{15}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{15}{2}x+\frac{225}{16}=25+\frac{225}{16}$$

Add 25 to \frac{225}{16}.

$$x^{2}+\frac{15}{2}x+\frac{225}{16}=\frac{625}{16}$$

Factor x^{2}+\frac{15}{2}x+\frac{225}{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}.

$$\left(x+\frac{15}{4}\right)^{2}=\frac{625}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{15}{4}\right)^{2}}=\sqrt{\frac{625}{16}}$$

Simplify.

$$x+\frac{15}{4}=\frac{25}{4} x+\frac{15}{4}=-\frac{25}{4}$$

Subtract \frac{15}{4} from both sides of the equation.

$$x=\frac{5}{2} x=-10$$