We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.
 
+0  
 
0
151
3
avatar

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
avatar
0

xxxxxx.

 Feb 4, 2019
edited by Guest  Feb 4, 2019
 #3
avatar+104830 
+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

21 Online Users

avatar