Maths Olympiad Prep

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Algebra Difficulty 5.4 AIME, harder Prove it Bulgaria

Given the equation
2+x9+62=x5−26+6−23+2. 2 + x\sqrt{9 + 6\sqrt{2}} = x\sqrt{5 - 2\sqrt{6}} + \sqrt{6} - 2\sqrt{3} + \sqrt{2}.
a) Write the root of the equation in the form m−nm - \sqrt{n}, where mm and nn are natural numbers.
b) Factor the expression a3−3a2−5a+7a^3 - 3a^2 - 5a + 7 into two non-constant factors with integer coefficients and calculate the value of this expression if aa is the root found in a).

Solution

9+62=32+22+1=3(2+1)=6+3 \sqrt{9 + 6\sqrt{2}} = \sqrt{3}\sqrt{2 + 2\sqrt{2} + 1} = \sqrt{3}(\sqrt{2} + 1) = \sqrt{6} + \sqrt{3}
5−26=3−26+2=∣3−2∣=3−2. \sqrt{5 - 2\sqrt{6}} = \sqrt{3 - 2\sqrt{6} + 2} = |\sqrt{3} - \sqrt{2}| = \sqrt{3} - \sqrt{2}.
The equation takes the form
x(6+3−3+2)=6+2−23−2 x(\sqrt{6} + \sqrt{3} - \sqrt{3} + \sqrt{2}) = \sqrt{6} + \sqrt{2} - 2\sqrt{3} - 2
x(6+2)=(6+2)(1−2), x(\sqrt{6} + \sqrt{2}) = (\sqrt{6} + \sqrt{2})(1 - \sqrt{2}),
whence (given 6+2>0\sqrt{6} + \sqrt{2} > 0) finally x=1−2x = 1 - \sqrt{2}.

b) We have a3−3a2−5a+7=a3−a2−2a2+2a−7a+7=(a−1)(a2−2a−7)a^3 - 3a^2 - 5a + 7 = a^3 - a^2 - 2a^2 + 2a - 7a + 7 = (a-1)(a^2 - 2a - 7). The product of a−1=−2a-1 = -\sqrt{2} and a2−2a−7=(a−1)2−8=2−8=−6a^2 - 2a - 7 = (a-1)^2 - 8 = 2 - 8 = -6 is 626\sqrt{2}.
□\square

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