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

Sep 9, 2018

#1
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Solve for a:
5 sqrt(4 a^2 + 1) = 4 a^2 + 7

Raise both sides to the power of two:
25 (4 a^2 + 1) = (4 a^2 + 7)^2

Expand out terms of the left hand side:
100 a^2 + 25 = (4 a^2 + 7)^2

Expand out terms of the right hand side:
100 a^2 + 25 = 16 a^4 + 56 a^2 + 49

Subtract 16 a^4 + 56 a^2 + 49 from both sides:
-16 a^4 + 44 a^2 - 24 = 0

Substitute x = a^2:
-16 x^2 + 44 x - 24 = 0

The left hand side factors into a product with three terms:
-4 (x - 2) (4 x - 3) = 0

Divide both sides by -4:
(x - 2) (4 x - 3) = 0

Split into two equations:
x - 2 = 0 or 4 x - 3 = 0

x = 2 or 4 x - 3 = 0

Substitute back for x = a^2:
a^2 = 2 or 4 x - 3 = 0

Take the square root of both sides:
a = sqrt(2) or a = -sqrt(2) or 4 x - 3 = 0

a = sqrt(2) or a = -sqrt(2) or 4 x = 3

Divide both sides by 4:
a = sqrt(2) or a = -sqrt(2) or x = 3/4

Substitute back for x = a^2:
a = sqrt(2) or a = -sqrt(2) or a^2 = 3/4

Take the square root of both sides:
a = sqrt(2) is the largest value that satisfies the equation.
a = sqrt(2)    or    a = -sqrt(2)    or    a = sqrt(3)/2    or    a = -sqrt(3)/2

Sep 9, 2018
edited by Guest  Sep 9, 2018
#2
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$$\frac{7\sqrt{(2a)^2+(1)^2}-4a^2-1}{\sqrt{1+4a^2}+3}\,=\,2$$

Multiply both sides of the equation by   $$\sqrt{1+4a^2}+3$$   and note that  $$\sqrt{1+4a^2}+3\neq0\\~\\ \sqrt{1+4a^2}\neq-3$$

But there isn't a value of  a  that makes that true.

$$7\sqrt{(2a)^2+(1)^2}-4a^2-1\,=2(\sqrt{1+4a^2}+3)\\~\\ 7\sqrt{4a^2+1}-4a^2-1\,=\,2\sqrt{1+4a^2}+6\\~\\ 7\sqrt{4a^2+1}\,=\,2\sqrt{4a^2+1}+7+4a^2\\~\\ 7\sqrt{4a^2+1}-2\sqrt{4a^2+1}\,=\,7+4a^2\\~\\ 5\sqrt{4a^2+1}\,=\,7+4a^2\\~\\ (5\sqrt{4a^2+1})^2\,=\,(7+4a^2)^2\\~\\ 25(4a^2+1)\,=\,(7+4a^2)(7+4a^2)\\~\\ 100a^2+25\,=\,49+56a^2+16a^4\\~\\ -16a^4+44a^2-24\,=\,0\\~\\ -16(a^2)^2+44(a^2)-24\,=\,0$$

Let  a2  =  u  , and let's substitute   u  in for  a2 .

$$-16u^2+44u-24\,=\,0\\~\\ -4u^2+11u-6\,=\,0\\~\\ -4u^2+8u+3u-6\,=\,0\\~\\ -4u(u-2)+3(u-2)\,=\,0\\~\\ (u-2)(-4u+3)\,=\,0\\~\\ u-2=0\qquad\text{or}\qquad-4u+3=0\\~\\ \phantom{--}u=2\qquad\text{or}\qquad u=\frac34$$

Now substitute  a2  back in for  u .

$$\begin{array}{ccc} a^2=2&\qquad\text{or}\qquad&a^2=\frac34\\~\\ a=\pm\sqrt2&\qquad\text{or}\qquad&a=\pm\sqrt{\frac34}\\~\\ &&a=\pm\frac{\sqrt3}{2}\\~\\ a=\sqrt2\qquad\text{or}\qquad a=-\sqrt2&\text{or}&a=\frac{\sqrt3}{2}\qquad\text{or}\qquad a=-\frac{\sqrt3}{2} \end{array}$$

The greatest value of  a  that satisfies the equation is  $$\sqrt2$$  .

Sep 9, 2018
edited by hectictar  Sep 9, 2018
edited by hectictar  Sep 9, 2018
#3
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a=sqrt(2)

5 sqrt(4 a^2 + 1) = 4 a^2 + 7    sub sqrt(2) for a:

5 sqrt(4sqrt(2)^2 +1) = 4sqrt(2)^2 + 7

5sqrt(4*2  + 1) = 4 * 2 + 7

5 * 3  = 8 + 7

15  =  15

So, the sqrt(2) is the largest "a" that satisfies the equation.

Sep 9, 2018