Problem of the Week

Updated at Mar 25, 2024 5:43 PM

This week we have another equation problem:

How would you solve the equation \({(4p)}^{2}+\frac{3}{2+p}=17\)?

Let's start!



\[{(4p)}^{2}+\frac{3}{2+p}=17\]

1
Use Multiplication Distributive Property: \({(xy)}^{a}={x}^{a}{y}^{a}\).
\[{4}^{2}{p}^{2}+\frac{3}{2+p}=17\]

2
Simplify  \({4}^{2}\)  to  \(16\).
\[16{p}^{2}+\frac{3}{2+p}=17\]

3
Multiply both sides by \(2+p\).
\[16{p}^{2}(2+p)+3=17(2+p)\]

4
Simplify.
\[32{p}^{2}+16{p}^{3}+3=34+17p\]

5
Move all terms to one side.
\[32{p}^{2}+16{p}^{3}+3-34-17p=0\]

6
Simplify  \(32{p}^{2}+16{p}^{3}+3-34-17p\)  to  \(32{p}^{2}+16{p}^{3}-31-17p\).
\[32{p}^{2}+16{p}^{3}-31-17p=0\]

7
Factor \(32{p}^{2}+16{p}^{3}-31-17p\) using Polynomial Division.
\[(16{p}^{2}+48p+31)(p-1)=0\]

8
Solve for \(p\).
\[p=1\]

9
Use the Quadratic Formula.
\[p=\frac{-48+8\sqrt{5}}{32},\frac{-48-8\sqrt{5}}{32}\]

10
Collect all solutions from the previous steps.
\[p=1,\frac{-48+8\sqrt{5}}{32},\frac{-48-8\sqrt{5}}{32}\]

11
Simplify solutions.
\[p=1,-\frac{6-\sqrt{5}}{4},-\frac{6+\sqrt{5}}{4}\]

Done

Decimal Form: 1, -0.940983, -2.059017
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