Problem of the Week

Updated at Oct 21, 2024 4:02 PM

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Mar 31, 2025

How would you solve $(2+{n}^{2})\times \frac{2+n}{3}=63$ ?

Below is the solution.

(2+{n}^{2})\times \frac{2+n}{3}=63

Use this rule:

a \times \frac{b}{c}=\frac{ab}{c}

\frac{(2+{n}^{2})(2+n)}{3}=63

Multiply both sides by

3

(2+{n}^{2})(2+n)=189

Expand.

4+2n+2{n}^{2}+{n}^{3}=189

Move all terms to one side.

4+2n+2{n}^{2}+{n}^{3}-189=0

Simplify

4+2n+2{n}^{2}+{n}^{3}-189

-185+2n+2{n}^{2}+{n}^{3}

-185+2n+2{n}^{2}+{n}^{3}=0

Factor

-185+2n+2{n}^{2}+{n}^{3}

using Polynomial Division.

Factor the following.

-185+2n+2{n}^{2}+{n}^{3}

First, find all factors of the constant term 185.

1, 5, 37, 185

Try each factor above using the Remainder Theorem.

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

-185+2\times 1+2\times {1}^{2}+{1}^{3} = -180

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

-185+2\times -1+2{(-1)}^{2}+{(-1)}^{3} = -186

Substitute 5 into n. Since the result is 0, n-5 is a factor..

-185+2\times 5+2\times {5}^{2}+{5}^{3} = 0

n-5

Polynomial Division: Divide

-185+2n+2{n}^{2}+{n}^{3}

n-5

Rewrite the expression using the above.

{n}^{2}+7n+37

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({n}^{2}+7n+37)(n-5)=0

Solve for

n

n=5

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=7

and

c=37

{n}^{}=\frac{-7+\sqrt{{7}^{2}-4\times 37}}{2},\frac{-7-\sqrt{{7}^{2}-4\times 37}}{2}

Simplify.

n=\frac{-7+3\sqrt{11}\imath }{2},\frac{-7-3\sqrt{11}\imath }{2}

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n=\frac{-7+3\sqrt{11}\imath }{2},\frac{-7-3\sqrt{11}\imath }{2}

Collect all solutions from the previous steps.

n=5,\frac{-7+3\sqrt{11}\imath }{2},\frac{-7-3\sqrt{11}\imath }{2}

Done