Problem of the Week

Updated at Jul 19, 2021 4:12 PM

How can we solve the equation (5n)2×3−n2=25{(\frac{5}{n})}^{2}\times \frac{3-n}{2}=25?

Below is the solution.



(5n)2×3−n2=25{(\frac{5}{n})}^{2}\times \frac{3-n}{2}=25

1
Use Division Distributive Property: (xy)a=xaya{(\frac{x}{y})}^{a}=\frac{{x}^{a}}{{y}^{a}}.
52n2×3−n2=25\frac{{5}^{2}}{{n}^{2}}\times \frac{3-n}{2}=25

2
Simplify  52{5}^{2}  to  2525.
25n2×3−n2=25\frac{25}{{n}^{2}}\times \frac{3-n}{2}=25

3
Use this rule: ab×cd=acbd\frac{a}{b} \times \frac{c}{d}=\frac{ac}{bd}.
25(3−n)n2×2=25\frac{25(3-n)}{{n}^{2}\times 2}=25

4
Regroup terms.
25(3−n)2n2=25\frac{25(3-n)}{2{n}^{2}}=25

5
Multiply both sides by 2n22{n}^{2}.
25(3−n)=25×2n225(3-n)=25\times 2{n}^{2}

6
Simplify  25×2n225\times 2{n}^{2}  to  50n250{n}^{2}.
25(3−n)=50n225(3-n)=50{n}^{2}

7
Divide both sides by 2525.
3−n=2n23-n=2{n}^{2}

8
Move all terms to one side.
3−n−2n2=03-n-2{n}^{2}=0

9
Multiply both sides by −1-1.
2n2+n−3=02{n}^{2}+n-3=0

10
Split the second term in 2n2+n−32{n}^{2}+n-3 into two terms.
2n2+3n−2n−3=02{n}^{2}+3n-2n-3=0

11
Factor out common terms in the first two terms, then in the last two terms.
n(2n+3)−(2n+3)=0n(2n+3)-(2n+3)=0

12
Factor out the common term 2n+32n+3.
(2n+3)(n−1)=0(2n+3)(n-1)=0

13
Solve for nn.
n=−32,1n=-\frac{3}{2},1

Done

Decimal Form: -1.5, 1
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