Problem of the Week

Updated at Jan 8, 2024 8:46 AM

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

How would you solve the equation \(\frac{3}{(2+w){(w-3)}^{2}}=\frac{1}{4}\)?

Here are the steps:



\[\frac{3}{(2+w){(w-3)}^{2}}=\frac{1}{4}\]

1
Multiply both sides by \((2+w){(w-3)}^{2}\).
\[3=\frac{1}{4}(2+w){(w-3)}^{2}\]

2
Simplify  \(\frac{1}{4}(2+w){(w-3)}^{2}\)  to  \(\frac{(2+w){(w-3)}^{2}}{4}\).
\[3=\frac{(2+w){(w-3)}^{2}}{4}\]

3
Multiply both sides by \(4\).
\[12=(2+w){(w-3)}^{2}\]

4
Expand.
\[12=2{w}^{2}-12w+18+{w}^{3}-6{w}^{2}+9w\]

5
Simplify  \(2{w}^{2}-12w+18+{w}^{3}-6{w}^{2}+9w\)  to  \(-4{w}^{2}-3w+18+{w}^{3}\).
\[12=-4{w}^{2}-3w+18+{w}^{3}\]

6
Move all terms to one side.
\[12+4{w}^{2}+3w-18-{w}^{3}=0\]

7
Simplify  \(12+4{w}^{2}+3w-18-{w}^{3}\)  to  \(-6+4{w}^{2}+3w-{w}^{3}\).
\[-6+4{w}^{2}+3w-{w}^{3}=0\]

8
Factor \(-6+4{w}^{2}+3w-{w}^{3}\) using Polynomial Division.
\[(-{w}^{2}+3w+6)(w-1)=0\]

9
Solve for \(w\).
\[w=1\]

10
Use the Quadratic Formula.
\[w=\frac{-3+\sqrt{33}}{-2},\frac{-3-\sqrt{33}}{-2}\]

11
Collect all solutions from the previous steps.
\[w=1,\frac{-3+\sqrt{33}}{-2},\frac{-3-\sqrt{33}}{-2}\]

12
Simplify solutions.
\[w=1,-\frac{-3+\sqrt{33}}{2},-\frac{-3-\sqrt{33}}{2}\]

Done

Decimal Form: 1, -1.372281, 4.372281
Cymath foriOSAndroid