Solve 3/x^2-5/x-2=0

$$\frac { 3 } { x ^ { 2 } } - \frac { 5 } { x } - 2 = 0$$

Solve for x

$$x=-3$$

$$x=\frac{1}{2}=0.5$$

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 x^{2}, the least common multiple of x^{2},x.

$$3-x\times 5+x^{2}\left(-2\right)=0$$

Multiply -1 and 5 to get -5.

$$3-5x+x^{2}\left(-2\right)=0$$

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

$$-2x^{2}-5x+3=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -2x^{2}+ax+bx+3. To find a and b, set up a system to be solved.

$$a+b=-5 ab=-2\times 3=-6$$

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 -6.

$$1,-6 2,-3$$

Calculate the sum for each pair.

$$1-6=-5 2-3=-1$$

The solution is the pair that gives sum -5.

$$a=1 b=-6$$

Rewrite -2x^{2}-5x+3 as \left(-2x^{2}+x\right)+\left(-6x+3\right).

$$\left(-2x^{2}+x\right)+\left(-6x+3\right)$$

Factor out -x in the first and -3 in the second group.

$$-x\left(2x-1\right)-3\left(2x-1\right)$$

Factor out common term 2x-1 by using distributive property.

$$\left(2x-1\right)\left(-x-3\right)$$

To find equation solutions, solve 2x-1=0 and -x-3=0.

$$x=\frac{1}{2} x=-3$$

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 x^{2}, the least common multiple of x^{2},x.

$$3-x\times 5+x^{2}\left(-2\right)=0$$

Multiply -1 and 5 to get -5.

$$3-5x+x^{2}\left(-2\right)=0$$

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.

$$-2x^{2}-5x+3=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -2 for a, -5 for b, and 3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\left(-2\right)\times 3}}{2\left(-2\right)}$$

Square -5.

$$x=\frac{-\left(-5\right)±\sqrt{25-4\left(-2\right)\times 3}}{2\left(-2\right)}$$

Multiply -4 times -2.

$$x=\frac{-\left(-5\right)±\sqrt{25+8\times 3}}{2\left(-2\right)}$$

Multiply 8 times 3.

$$x=\frac{-\left(-5\right)±\sqrt{25+24}}{2\left(-2\right)}$$

Add 25 to 24.

$$x=\frac{-\left(-5\right)±\sqrt{49}}{2\left(-2\right)}$$

Take the square root of 49.

$$x=\frac{-\left(-5\right)±7}{2\left(-2\right)}$$

The opposite of -5 is 5.

$$x=\frac{5±7}{2\left(-2\right)}$$

Multiply 2 times -2.

$$x=\frac{5±7}{-4}$$

Now solve the equation x=\frac{5±7}{-4} when ± is plus. Add 5 to 7.

$$x=\frac{12}{-4}$$

Divide 12 by -4.

$$x=-3$$

Now solve the equation x=\frac{5±7}{-4} when ± is minus. Subtract 7 from 5.

$$x=-\frac{2}{-4}$$

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

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

The equation is now solved.

$$x=-3 x=\frac{1}{2}$$

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 x^{2}, the least common multiple of x^{2},x.

$$3-x\times 5+x^{2}\left(-2\right)=0$$

Subtract 3 from both sides. Anything subtracted from zero gives its negation.

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

Multiply -1 and 5 to get -5.

$$-5x+x^{2}\left(-2\right)=-3$$

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.

$$-2x^{2}-5x=-3$$

Divide both sides by -2.

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

Dividing by -2 undoes the multiplication by -2.

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

Divide -5 by -2.

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

Divide -3 by -2.

$$x^{2}+\frac{5}{2}x=\frac{3}{2}$$

Divide \frac{5}{2}, the coefficient of the x term, by 2 to get \frac{5}{4}. Then add the square of \frac{5}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{5}{2}x+\left(\frac{5}{4}\right)^{2}=\frac{3}{2}+\left(\frac{5}{4}\right)^{2}$$

Square \frac{5}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{5}{2}x+\frac{25}{16}=\frac{3}{2}+\frac{25}{16}$$

Add \frac{3}{2} to \frac{25}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{5}{2}x+\frac{25}{16}=\frac{49}{16}$$

Factor x^{2}+\frac{5}{2}x+\frac{25}{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}.

$$\left(x+\frac{5}{4}\right)^{2}=\frac{49}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{5}{4}\right)^{2}}=\sqrt{\frac{49}{16}}$$

Simplify.

$$x+\frac{5}{4}=\frac{7}{4} x+\frac{5}{4}=-\frac{7}{4}$$

Subtract \frac{5}{4} from both sides of the equation.

$$x=\frac{1}{2} x=-3$$