Sub-Problem 2

Revision as of 08:37, 23 March 2021 by Timz2005 (talk | contribs) (Video Solution)

Problem

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

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

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(Note: If anyone can code the "content" of this page I would be appreciated! I completely don't know how to code the "content" of a page on AoPS)

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=C180TL1PLaA

~North America Math Contest Go Go Go