\(\frac{7\sqrt{(2a)^2+(1)^2}-4a^2-1}{\sqrt{1+4a^2}+3}\,=\,2 \)
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 \(\sqrt2\) .
