Problem of the Week

Updated at Dec 15, 2025 10:11 AM

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

How would you solve \(\frac{2+4v}{{(\frac{5}{v})}^{2}}=22\)?

Here are the steps:



\[\frac{2+4v}{{(\frac{5}{v})}^{2}}=22\]

1
Factor out the common term \(2\).
\[\frac{2(1+2v)}{{(\frac{5}{v})}^{2}}=22\]

2
Use Division Distributive Property: \({(\frac{x}{y})}^{a}=\frac{{x}^{a}}{{y}^{a}}\).
\[\frac{2(1+2v)}{\frac{{5}^{2}}{{v}^{2}}}=22\]

3
Simplify  \({5}^{2}\)  to  \(25\).
\[\frac{2(1+2v)}{\frac{25}{{v}^{2}}}=22\]

4
Invert and multiply.
\[2(1+2v)\times \frac{{v}^{2}}{25}=22\]

5
Simplify  \(2(1+2v)\times \frac{{v}^{2}}{25}\)  to  \(\frac{2(1+2v){v}^{2}}{25}\).
\[\frac{2(1+2v){v}^{2}}{25}=22\]

6
Regroup terms.
\[\frac{2{v}^{2}(1+2v)}{25}=22\]

7
Multiply both sides by \(25\).
\[2{v}^{2}(1+2v)=550\]

8
Expand.
\[2{v}^{2}+4{v}^{3}=550\]

9
Move all terms to one side.
\[2{v}^{2}+4{v}^{3}-550=0\]

10
Factor out the common term \(2\).
\[2({v}^{2}+2{v}^{3}-275)=0\]

11
Factor \({v}^{2}+2{v}^{3}-275\) using Polynomial Division.
\[2(2{v}^{2}+11v+55)(v-5)=0\]

12
Solve for \(v\).
\[v=5\]

13
Use the Quadratic Formula.
\[v=\frac{-11+\sqrt{319}\imath }{4},\frac{-11-\sqrt{319}\imath }{4}\]

14
Collect all solutions from the previous steps.
\[v=5,\frac{-11+\sqrt{319}\imath }{4},\frac{-11-\sqrt{319}\imath }{4}\]

Done
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