Find the greatest a such that \(\frac{7\sqrt{(2a)^2+(1)^2}-4a^2-1}{\sqrt{1+4a^2}+3}=2\).

Guest Feb 3, 2019

#1**0 **

Getting rid of the fraction helps:

\(7\sqrt{\left(2a\right)^2+1}\:-\:4a^2-1=2\sqrt{1+4a^2}+6\)

Then, we subtract 6 from both sides.

Doing so, we have all of the contants on one side. Now, square the equation:

\(\left(-4a^2+7\sqrt{\left(2a\right)^2+1}-7\right)^2=\left(2\sqrt{1+4a^2}\right)^2 \rightarrow \\ 16a^4+252a^2-56a^2\sqrt{4a^2+1}-98\sqrt{4a^2+1}+98 =4+16a^2\)

Subtract 16a^4 and 252a^2 and 98 from both sides, and simplify:

\(-56a^2\sqrt{4a^2+1}-98\sqrt{4a^2+1}=-16a^4-236a^2-94\)

\(\mathrm{Subtract\:}-16a^4-236a^2-94\mathrm{\:from\:both\:sides}\)

You'll get the following:

\(-56a^2\sqrt{4a^2+1}-98\sqrt{4a^2+1}+16a^4+236a^2+94=0\)

Then factoring yields:

\(2\left(-28a^2\sqrt{4a^2+1}-49\sqrt{4a^2+1}+8a^4+118a^2+47\right)\)

Thus, this equation equals 0, and you can divide by 2. Now we subtract \(8a^4+118a^2\) and 47 from both sides.

Factoring again, \(-7\sqrt{4a^2+1}\left(4a^2+7\right)=-8a^4-118a^2-47\)

If you expand these, you will get a sextic equation (equation with degree six variables).

Plugging those into Wolfram yields \(a=\pm\sqrt 2, a=\pm\frac{\sqrt 3}{2}, \text{and } a=\pm \frac{\sqrt 2\sqrt{67+9\sqrt{57}}}{4}.\)

However, when you plug the last possibility in, it doesn't work. So, your answer is **sqrt 2. **

itsyaboi Feb 4, 2019

#3**+1 **

Rearrange as

7 √ [ (2a)^2 + (1)^2 ] - 4a^2 - 1 = 2 √[4a^2 + 1 ] + 6 rearrange again

7√ [ 4a^2 + 1 ] = 2 √[4a^2 + 1 ] + 6 + [4a^2 + 1] rearrange again

[ 4a^2 + 1 ] - 5√ [4a^2 + 1 ] + 6 = 0

Now....Let [4a^2 + 1 ] = m^2 ⇒ √ [4a^2 + 1 ] = m ...... and we have that

m^2 - 5m + 6 = 0 factor

(m -3) (m - 2) = 0

Setting each factor to 0 an solve for m and we get that

m = 2 or m = 3

So

m^2 = 4 or m^2 = 9

If m^2 = 4 then we have that

4a^2 + 1 = 4

4a^2 = 3

a^2 = 3/4

a = ± √[3] /2

If m = 9,then we have that

4a^2 + 1 = 9

4a^2 = 8

a^2 =8/4

a^2 = 2

a = ± √2

So.....the largest value for a is √2

CPhill Feb 4, 2019