Difference between revisions of "Sub-Problem 2"

Latest revision as of 21:23, 28 November 2023

(b) Determine all $(a,b)$ such that: $\sqrt{a} + \sqrt{b} = 8$ $\log_{10} a + \log_{10} (b) = 2$

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 +- 8\sqrt(6)

Since a = 44 - b, two solutions are:

$(a,b) = (22 + 8\sqrt6, 22 - 8\sqrt6)$ $(a,b) = (22 - 8\sqrt6, 22 + 8\sqrt6)$

~North America Math Contest Go Go Go

~North America Math Contest Go Go Go

@@ Line 33: / Line 33: @@
 Since a = 44 - b, two solutions are:
-<cmath>(a,b) = (22 + 8\sqrt(6), 22 - 8\sqrt6))</cmath>
+<cmath>(a,b) = (22 + 8\sqrt6, 22 - 8\sqrt6)</cmath>
-<cmath>(a,b) = (22 - 8\sqrt(6), 22 + 8\sqrt(6))</cmath>
+<cmath>(a,b) = (22 - 8\sqrt6, 22 + 8\sqrt6)</cmath>
 ~North America Math Contest Go Go Go