Submitted by Grainne on Sun, 09/10/2017 - 10:01
Forums:
in the hints and partial solutions for question 3iia. it is stated that there is only one solution to this question, however when I completed the question I ended up with x=2 and x=142. I plugged both into the original equation and x=2 works and when plugging in x=142 if the negative square root of 144 is taken instead of the positive, this solution also works. Have I made a mistake?
You haven't made a mistake
You haven't made a mistake per se but bear in mind the radical sign $\sqrt{\,\cdot\,}$ necessarily means the positive square root of its argument so you can't "choose the negative root" (unless you multiply it by $-1$). Hence, $x=142$ is not a solution of the equation.