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Difference between revisions of "Mock AIME 1 Pre 2005 Problems/Problem 5"

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== Solution ==
 
== Solution ==
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Let <math>a=\sqrt[3]{x}, b = \sqrt[3]{20-x}</math>. Then <math>a+b = 2</math> and <math>a^3 + b^3 = 20</math>. Factoring, <math>a^3 + b^3 = (a+b)((a+b)^2-3ab) = 2(4-3ab)= 8-6ab=20 \Longrightarrow ab = -2</math>.
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Let <math>a=\sqrt[3]{x}, b = \sqrt[3]{20-x}</math>. Then <math>a+b = 2</math> and <math>a^3 + b^3 = 20</math>. Factoring, <cmath>a^3 + b^3 = (a+b)((a+b)^2-3ab) = 2(4-3ab)= 8-6ab=20 \Longrightarrow ab = -2</cmath>  
  
 
Solving <math>a+b=2, ab=-2</math> gives us the quadratic <math>a^2 - 2a - 2 = 0</math>. The [[quadratic formula]] yields <math>a = \frac{2 - \sqrt{12}}{2} = 1 - \sqrt{3}</math>, and <math>x = a^3 = (1-\sqrt{3})^3 = 1 - 3\sqrt{3} + 9 - 3\sqrt{3} = 10 - \sqrt{108}</math>. Therefore, <math>p+q=\boxed{118}</math>.
 
Solving <math>a+b=2, ab=-2</math> gives us the quadratic <math>a^2 - 2a - 2 = 0</math>. The [[quadratic formula]] yields <math>a = \frac{2 - \sqrt{12}}{2} = 1 - \sqrt{3}</math>, and <math>x = a^3 = (1-\sqrt{3})^3 = 1 - 3\sqrt{3} + 9 - 3\sqrt{3} = 10 - \sqrt{108}</math>. Therefore, <math>p+q=\boxed{118}</math>.

Latest revision as of 16:46, 21 March 2008

Problem

Let $a$ and $b$ be the two real values of $x$ for which \[\sqrt[3]{x} + \sqrt[3]{20 - x} = 2\] The smaller of the two values can be expressed as $p - \sqrt{q}$, where $p$ and $q$ are integers. Compute $p + q$.

Solution

Let $a=\sqrt[3]{x}, b = \sqrt[3]{20-x}$. Then $a+b = 2$ and $a^3 + b^3 = 20$. Factoring, \[a^3 + b^3 = (a+b)((a+b)^2-3ab) = 2(4-3ab)= 8-6ab=20 \Longrightarrow ab = -2\]

Solving $a+b=2, ab=-2$ gives us the quadratic $a^2 - 2a - 2 = 0$. The quadratic formula yields $a = \frac{2 - \sqrt{12}}{2} = 1 - \sqrt{3}$, and $x = a^3 = (1-\sqrt{3})^3 = 1 - 3\sqrt{3} + 9 - 3\sqrt{3} = 10 - \sqrt{108}$. Therefore, $p+q=\boxed{118}$.

See also

Mock AIME 1 Pre 2005 (Problems, Source)
Preceded by
Problem 4 		Followed by
Problem 6
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
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