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- Feb 14, 2012
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$a$ is a real number that satisfies the equation $a-\sqrt{4-a}\sqrt{5-a}=\sqrt{5-a}\sqrt{6-a}+\sqrt{6-a}\sqrt{4-a}$.
Find $a$.
Find $a$.
My solution$a$ is a real number that satisfies the equation $a-\sqrt{4-a}\sqrt{5-a}=\sqrt{5-a}\sqrt{6-a}+\sqrt{6-a}\sqrt{4-a}$.
Find $a$.
Well done, Jester! My hat is off to you for thinking out such a nice way to factorize the given expression!My solution
Moving everthing to one side of the equal sign, the expression can be factored as
$\left(\sqrt{4-a} + \sqrt{5-a}\right)\left(5\sqrt{4-a} - 4\sqrt{5-a} + \sqrt{6-a}\right) = 0.$
Cleary the first term cannot be zero so the second must. Re-arranging such that
$5\sqrt{4-a} + \sqrt{6-a} = 4\sqrt{5-a}$
and squaring both sides and isolating the square root term gives
$10\sqrt{4-a}\sqrt{6-a} = 10a - 26$
and squaring each side again and isolating $a$ give
$a = \dfrac{431}{120} = 3.591\dot{6}$
My solution:$a$ is a real number that satisfies the equation $a-\sqrt{4-a}\sqrt{5-a}=\sqrt{5-a}\sqrt{6-a}+\sqrt{6-a}\sqrt{4-a}$.
Find $a$.
$5-a=\dfrac{m}{2}$ | $4-a=\dfrac{m}{2}-1$ | $6-a=\dfrac{m}{2}+1$ |