Re: Find a
anemone said:
$a$ 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
[TABLE="class: grid, width: 500"]
[TR]
[TD]$5-a=\dfrac{m}{2}$[/TD]
[TD]$4-a=\dfrac{m}{2}-1$[/TD]
[TD]$6-a=\dfrac{m}{2}+1$[/TD]
[/TR]
[/TABLE]
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, $a=\dfrac{10-m}{2}=\dfrac{10-2(\dfrac{169}{120})}{2}=\dfrac{431}{120}$.