Problem of the Week

Updated at Sep 29, 2025 8:31 AM

For this week we've brought you this equation problem.

How would you solve the equation \(\frac{20}{u}+{(2+u)}^{2}=29\)?

Here are the steps:



\[\frac{20}{u}+{(2+u)}^{2}=29\]

1
Multiply both sides by \(u\).
\[20+u{(2+u)}^{2}=29u\]

2
Simplify.
\[20+4u+4{u}^{2}+{u}^{3}=29u\]

3
Move all terms to one side.
\[20+4u+4{u}^{2}+{u}^{3}-29u=0\]

4
Simplify  \(20+4u+4{u}^{2}+{u}^{3}-29u\)  to  \(20-25u+4{u}^{2}+{u}^{3}\).
\[20-25u+4{u}^{2}+{u}^{3}=0\]

5
Factor \(20-25u+4{u}^{2}+{u}^{3}\) using Polynomial Division.
\[({u}^{2}+5u-20)(u-1)=0\]

6
Solve for \(u\).
\[u=1\]

7
Use the Quadratic Formula.
\[u=\frac{-5+\sqrt{105}}{2},\frac{-5-\sqrt{105}}{2}\]

8
Collect all solutions from the previous steps.
\[u=1,\frac{-5+\sqrt{105}}{2},\frac{-5-\sqrt{105}}{2}\]

Done

Decimal Form: 1, 2.623475, -7.623475
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