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# help plz

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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
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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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xxxxxx.

Feb 4, 2019
edited by Guest  Feb 4, 2019
#3
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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

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