Difference between revisions of "Sub-Problem 2"
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== Problem == | == Problem == | ||
− | (b) Determine all (a,b) such that: | + | (b) Determine all <math>(a,b)</math> such that: |
− | + | <cmath>\sqrt{a} + \sqrt{b} = 8</cmath> <cmath>\log_{10} a + \log_{10} (b) = 2</cmath> | |
+ | |||
+ | == Solution 1 == | ||
+ | From equation 2, we can acquire ab = 100 | ||
+ | |||
+ | We can then expand both sides by squaring: | ||
+ | |||
+ | <cmath>(\sqrt{a} + \sqrt{b})^2 = (8)^2</cmath> | ||
+ | <cmath>(a + b + 2 \sqrt{ab}) = 64</cmath> | ||
+ | |||
+ | since ab = 100: 2root(ab) is 2root(100), which is 20. | ||
+ | |||
+ | We can get the below equation: | ||
+ | |||
+ | <cmath>(a + b) = 44</cmath> | ||
+ | <cmath>(ab) = 100</cmath> | ||
+ | |||
+ | Substitue b = 44 - a, we get | ||
+ | |||
+ | <cmath>((44-a)a) = 100</cmath> | ||
+ | <cmath>(44a - a^2 - 100) = 0</cmath> | ||
+ | |||
+ | By quadratic equations Formula: | ||
+ | |||
+ | <math>{a=\frac{-44 \pm \sqrt{44^2-4(-1)(-100)}}{2(-1)}}</math> | ||
+ | |||
+ | <math>{a=\frac{44 \pm \sqrt{1536}}{2(1)}}</math> | ||
+ | |||
+ | which leads to the answer of 22 +- 8\sqrt(6) | ||
+ | |||
+ | Since a = 44 - b, two solutions are: | ||
+ | |||
+ | <cmath>(a,b) = (22 + 8\sqrt6, 22 - 8\sqrt6)</cmath> | ||
+ | <cmath>(a,b) = (22 - 8\sqrt6, 22 + 8\sqrt6)</cmath> | ||
+ | |||
+ | ~North America Math Contest Go Go Go | ||
+ | |||
+ | == Video Solution == | ||
+ | https://www.youtube.com/watch?v=C180TL1PLaA | ||
+ | |||
+ | ~North America Math Contest Go Go Go |
Latest revision as of 21:23, 28 November 2023
Problem
(b) Determine all such that:
Solution 1
From equation 2, we can acquire ab = 100
We can then expand both sides by squaring:
since ab = 100: 2root(ab) is 2root(100), which is 20.
We can get the below equation:
Substitue b = 44 - a, we get
By quadratic equations Formula:
which leads to the answer of 22 +- 8\sqrt(6)
Since a = 44 - b, two solutions are:
~North America Math Contest Go Go Go
Video Solution
https://www.youtube.com/watch?v=C180TL1PLaA
~North America Math Contest Go Go Go