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

 Feb 3, 2019
 #1
avatar+79 
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. 

 Feb 4, 2019
 #2
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0

xxxxxx.

 Feb 4, 2019
edited by Guest  Feb 4, 2019
 #3
avatar+98153 
+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

 

 

cool cool cool

 Feb 4, 2019
edited by CPhill  Feb 4, 2019
edited by CPhill  Feb 4, 2019

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