Difference between revisions of "Sub-Problem 2"

(Solution 1)
(Solution 1)
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Since a = 44 - b, two solutions are:
 
Since a = 44 - b, two solutions are:
  
<cmath>(a,b) = (</cmath>(8 \sqrt{6) + 22),<math></math>(-8 \sqrt{6} + 22))
+
<cmath>(a,b) = (22 + 8root(6), 22 - 8root(6))</cmath>
 +
<cmath>(a,b) = (22 - 8root(6), 22 + 8root(6))</cmath>
 +
 
 +
~North America Math Contest Go Go Go
  
 
== Video Solution ==
 
== Video Solution ==

Revision as of 18:55, 22 March 2021

Problem

(b) Determine all (a,b) such that:

                  \[\sqrt{a} + \sqrt{b} = 8\] \[\log_{10} a +  \log_{10} (b) = 2\]

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Solution 1

From equation 2, we can acquire ab = 100

We can then expand both sides by squaring:

\[(\sqrt{a} + \sqrt{b})^2 = (8)^2\] \[(a + b + 2 \sqrt{ab}) = 64\]

since ab = 100: 2root(ab) is 2root(100), which is 20.

We can get the below equation:

\[(a + b) = 44\] \[(ab) = 100\]

Substitue b = 44 - a, we get

\[((44-a)a) = 100\] \[(44a - a^2 - 100) = 0\]

By quadratic equations Formula:

${a=\frac{-44 \pm \sqrt{44^2-4(-1)(-100)}}{2(-1)}}$

${a=\frac{44 \pm \sqrt{1536}}{2(1)}}$

which leads to the answer of 22 +- 8root(6)

Since a = 44 - b, two solutions are:

\[(a,b) = (22 + 8root(6), 22 - 8root(6))\] \[(a,b) = (22 - 8root(6), 22 + 8root(6))\]

~North America Math Contest Go Go Go

Video Solution

https://www.youtube.com/watch?v=uQzjgxEEQ74&list=PLexHyfQ8DMuK6iQRknnbzBDtvxiu4JLD-&index=2

~NAMCG