Problem of the Week

Updated at Nov 5, 2018 11:02 AM

Recent Posts

Mar 31, 2025

How would you solve $\frac{20}{x}-2-{x}^{2}=4$ ?

Below is the solution.

\frac{20}{x}-2-{x}^{2}=4

Multiply both sides by

x

20-2x-{x}^{3}=4x

Move all terms to one side.

20-2x-{x}^{3}-4x=0

Simplify

20-2x-{x}^{3}-4x

20-6x-{x}^{3}

20-6x-{x}^{3}=0

Factor

20-6x-{x}^{3}

using Polynomial Division.

Factor the following.

20-6x-{x}^{3}

First, find all factors of the constant term 20.

1, 2, 4, 5, 10, 20

Try each factor above using the Remainder Theorem.

Substitute 1 into x. Since the result is not 0, x-1 is not a factor..

20-6\times 1-{1}^{3} = 13

Substitute -1 into x. Since the result is not 0, x+1 is not a factor..

20-6\times -1-{(-1)}^{3} = 27

Substitute 2 into x. Since the result is 0, x-2 is a factor..

20-6\times 2-{2}^{3} = 0

x-2

Polynomial Division: Divide

20-6x-{x}^{3}

x-2

Rewrite the expression using the above.

-{x}^{2}-2x-10

To get access to all 'How?' and 'Why?' steps, join Cymath Plus!

(-{x}^{2}-2x-10)(x-2)=0

Solve for

x

x=2

Use the Quadratic Formula.

In general, given

a{x}^{2}+bx+c=0

, there exists two solutions where:

x=\frac{-b+\sqrt{{b}^{2}-4ac}}{2a},\frac{-b-\sqrt{{b}^{2}-4ac}}{2a}

In this case,

a=-1

b=-2

and

c=-10

{x}^{}=\frac{2+\sqrt{{(-2)}^{2}-4\times 10}}{-2},\frac{2-\sqrt{{(-2)}^{2}-4\times 10}}{-2}

Simplify.

x=\frac{2+6\imath }{-2},\frac{2-6\imath }{-2}

To get access to all 'How?' and 'Why?' steps, join Cymath Plus!

x=\frac{2+6\imath }{-2},\frac{2-6\imath }{-2}

Collect all solutions from the previous steps.

x=2,\frac{2+6\imath }{-2},\frac{2-6\imath }{-2}

Simplify solutions.

x=2,-1-3\imath ,-1+3\imath

Done