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Find "a"

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anemone

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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 $a$.
 

Jester

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MHB Math Helper
Jan 26, 2012
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Re: Find a

$a$ 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 $a$.
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

$a = \dfrac{431}{120} = 3.591\dot{6}$
 
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anemone

MHB POTW Director
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Feb 14, 2012
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Re: Find a

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

$a = \dfrac{431}{120} = 3.591\dot{6}$
Well done, Jester! My hat is off to you for thinking out such a nice way to factorize the given expression! (Clapping)
 
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anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,684
Re: Find a

$a$ 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 $a$.
My solution:

I first let $m=(6-a)+(4-a)=10-2a=2(5-a)$, this gives

$5-a=\dfrac{m}{2}$$4-a=\dfrac{m}{2}-1$$6-a=\dfrac{m}{2}+1$

So the given equation becomes

$5-\dfrac{m}{2}-\sqrt{\dfrac{m}{2}-1}\sqrt{\dfrac{m}{2}}=\sqrt{ \dfrac{m}{2}}\sqrt{\dfrac{m}{2}+1}+\sqrt{\dfrac{m}{2}+1}\sqrt{\dfrac{m}{2}-1}$

$5-\dfrac{m}{2}=\sqrt{ \dfrac{m}{2}}\left(\sqrt{\dfrac{m}{2}+1}+\sqrt{ \dfrac{m}{2}-1} \right)+\sqrt{\dfrac{m}{2}+1}\sqrt{\dfrac{m}{2}-1}$

I then let $\sqrt{\dfrac{m}{2}}=k$ so I get

$5-k^2=k\left(\sqrt{k^2+1}+\sqrt{k^2-1} \right)+\sqrt{(k^2+1)(k^2-1)}$

Expanding and simplifying the equation and solving it for $k^2$ we get

$5-k^2-\sqrt{k^4-1}=k\left(\sqrt{k^2+1}+\sqrt{k^2-1} \right)$

$25-10k^2+k^4+k^4-1-2(5-k^2)\sqrt{k^4-1}=k^2\left(k^2+1+k^2-1+2\sqrt{k^2+1}\sqrt{k^2-1} \right)$

$25-10k^2+k^4+k^4-1-10\sqrt{k^4-1}+2k^2\sqrt{k^4-1}=k^2\left(2k^2+2\sqrt{k^4-1} \right)$

$25-10k^2+2k^4-1-10\sqrt{k^4-1}+2k^2\sqrt{k^4-1}=2k^4+2k^2\sqrt{k^4-1}$

$24-10k^2-10\sqrt{k^4-1}=0$

$k^2=\dfrac{169}{120}$

and therefore $\dfrac{m}{2}=k^2=\dfrac{169}{120}$ hence, $a=\dfrac{10-m}{2}=\dfrac{10-2(\dfrac{169}{120})}{2}=\dfrac{431}{120}$.