Find "a" which is a real number

[anemone]

$a$ 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 $a$.

 

[Mathwebster]

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

Spoiler

Moving everything 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

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

 

[anemone]

Re: Find a

My solution

Spoiler

Moving everything 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)

 

[anemone]

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, $a=\dfrac{10-m}{2}=\dfrac{10-2(\dfrac{169}{120})}{2}=\dfrac{431}{120}$.

 

[CellCoree]

To find $a$, we can manipulate the given equation algebraically. First, we can simplify the square roots by multiplying them together, which gives us:

$a - \sqrt{(4-a)(5-a)} = \sqrt{(5-a)(6-a)} + \sqrt{(6-a)(4-a)}$

Next, we can expand the parentheses on both sides of the equation:

$a - \sqrt{20 - 9a + a^2} = \sqrt{30 - 11a + a^2} + \sqrt{24 - 10a + a^2}$

We can then combine like terms and isolate the variable $a$ on one side of the equation:

$a - \sqrt{a^2 + 11a - 20} = \sqrt{a^2 - 11a + 30} + \sqrt{a^2 - 10a + 24}$

$a - \sqrt{(a+16)(a-5)} = \sqrt{(a+6)(a-5)} + \sqrt{(a+6)(a-4)}$

Finally, we can square both sides of the equation to eliminate the square roots:

$(a - \sqrt{(a+16)(a-5)})^2 = (\sqrt{(a+6)(a-5)} + \sqrt{(a+6)(a-4)})^2$

$a^2 - 2a\sqrt{(a+16)(a-5)} + (a+16)(a-5) = (a+6)(a-5) + 2\sqrt{(a+6)(a-5)(a+6)(a-4)} + (a+6)(a-4)$

Simplifying further, we get:

$a^2 + 11a - 80 = 2\sqrt{(a+6)(a-5)(a+6)(a-4)}$

Now, we can square both sides again and rearrange the equation to get a quadratic equation in terms of $a$:

$4(a+6)(a-5)(a+6)(a-4) = (a^2 + 11a - 80)^2$

$4(a^2 + 2a - 30)(a^2 + 2a - 24) = (a^2 + 11a - 80)^2$

$4a^4