Problem of the Week

Updated at Nov 13, 2023 11:35 AM

This week's problem comes from the equation category.

How would you solve (5q)22(q+2)=518\frac{{(\frac{5}{q})}^{2}}{2(q+2)}=\frac{5}{18}?

Let's begin!



(5q)22(q+2)=518\frac{{(\frac{5}{q})}^{2}}{2(q+2)}=\frac{5}{18}

1
Use Division Distributive Property: (xy)a=xaya{(\frac{x}{y})}^{a}=\frac{{x}^{a}}{{y}^{a}}.
52q22(q+2)=518\frac{\frac{{5}^{2}}{{q}^{2}}}{2(q+2)}=\frac{5}{18}

2
Simplify  52{5}^{2}  to  2525.
25q22(q+2)=518\frac{\frac{25}{{q}^{2}}}{2(q+2)}=\frac{5}{18}

3
Simplify  25q22(q+2)\frac{\frac{25}{{q}^{2}}}{2(q+2)}  to  252q2(q+2)\frac{25}{2{q}^{2}(q+2)}.
252q2(q+2)=518\frac{25}{2{q}^{2}(q+2)}=\frac{5}{18}

4
Multiply both sides by 2q2(q+2)2{q}^{2}(q+2).
25=518×2q2(q+2)25=\frac{5}{18}\times 2{q}^{2}(q+2)

5
Use this rule: ab×cd=acbd\frac{a}{b} \times \frac{c}{d}=\frac{ac}{bd}.
25=5×2q2(q+2)1825=\frac{5\times 2{q}^{2}(q+2)}{18}

6
Simplify  5×2q2(q+2)5\times 2{q}^{2}(q+2)  to  10q2(q+2)10{q}^{2}(q+2).
25=10q2(q+2)1825=\frac{10{q}^{2}(q+2)}{18}

7
Simplify  10q2(q+2)18\frac{10{q}^{2}(q+2)}{18}  to  5q2(q+2)9\frac{5{q}^{2}(q+2)}{9}.
25=5q2(q+2)925=\frac{5{q}^{2}(q+2)}{9}

8
Multiply both sides by 99.
225=5q2(q+2)225=5{q}^{2}(q+2)

9
Expand.
225=5q3+10q2225=5{q}^{3}+10{q}^{2}

10
Move all terms to one side.
225−5q3−10q2=0225-5{q}^{3}-10{q}^{2}=0

11
Factor out the common term 55.
5(45−q3−2q2)=05(45-{q}^{3}-2{q}^{2})=0

12
Factor 45−q3−2q245-{q}^{3}-2{q}^{2} using Polynomial Division.
5(−q2−5q−15)(q−3)=05(-{q}^{2}-5q-15)(q-3)=0

13
Solve for qq.
q=3q=3

14
Use the Quadratic Formula.
q=5+35ı−2,5−35ı−2q=\frac{5+\sqrt{35}\imath }{-2},\frac{5-\sqrt{35}\imath }{-2}

15
Collect all solutions from the previous steps.
q=3,5+35ı−2,5−35ı−2q=3,\frac{5+\sqrt{35}\imath }{-2},\frac{5-\sqrt{35}\imath }{-2}

16
Simplify solutions.
q=3,−5+35ı2,−5−35ı2q=3,-\frac{5+\sqrt{35}\imath }{2},-\frac{5-\sqrt{35}\imath }{2}

Done
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