Problem of the Week

Updated at Jul 15, 2024 2:53 PM

This week we have another equation problem:

How would you solve 4q+2(q−3)2=10\frac{4q+2}{{(q-3)}^{2}}=10?

Let's start!



4q+2(q−3)2=10\frac{4q+2}{{(q-3)}^{2}}=10

1
Factor out the common term 22.
2(2q+1)(q−3)2=10\frac{2(2q+1)}{{(q-3)}^{2}}=10

2
Multiply both sides by (q−3)2{(q-3)}^{2}.
2(2q+1)=10(q−3)22(2q+1)=10{(q-3)}^{2}

3
Divide both sides by 22.
2q+1=5(q−3)22q+1=5{(q-3)}^{2}

4
Expand.
2q+1=5q2−30q+452q+1=5{q}^{2}-30q+45

5
Move all terms to one side.
2q+1−5q2+30q−45=02q+1-5{q}^{2}+30q-45=0

6
Simplify  2q+1−5q2+30q−452q+1-5{q}^{2}+30q-45  to  32q−44−5q232q-44-5{q}^{2}.
32q−44−5q2=032q-44-5{q}^{2}=0

7
Multiply both sides by −1-1.
5q2−32q+44=05{q}^{2}-32q+44=0

8
Split the second term in 5q2−32q+445{q}^{2}-32q+44 into two terms.
5q2−10q−22q+44=05{q}^{2}-10q-22q+44=0

9
Factor out common terms in the first two terms, then in the last two terms.
5q(q−2)−22(q−2)=05q(q-2)-22(q-2)=0

10
Factor out the common term q−2q-2.
(q−2)(5q−22)=0(q-2)(5q-22)=0

11
Solve for qq.
q=2,225q=2,\frac{22}{5}

Done

Decimal Form: 2, 4.4
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