Hello, i was preparing for a test i have tomorrow so was doing review questions in my text book. I came across this one question where i knew what to do but got the wrong answer, must be a mistake in my working out

Question: the distance between (2,-7) and (a,-2) is $\displaystyle \sqrt{41}$units. Find the value of a if it is positive.

Edit: New to using Latex - sorry for any inconvenience

AND

The answer i have been obtaining is a=4

The formula to use was d = $\displaystyle \sqrt(x2-x1)^2+(y2-y1)^2$

in this case d being $\displaystyle \sqrt 41$

the actual answer is a= 6..

$\displaystyle \sqrt41 = \sqrt{(a-2)^2+(-2+7)^2}

\Rightarrow \sqrt41 = \sqrt{(a^2-4a+29)}

\Rightarrow 41 = a^2-4a+29

$

$\displaystyle \Rightarrow 12 = a^2-4a

\Rightarrow 12/a = a-4

$

$\displaystyle

\Rightarrow 12/a+4 = a

\Rightarrow 12+4 = a^2

$

$\displaystyle

\Rightarrow 16= a^2

\Rightarrow a=4

$