skip to content

STEP Support Programme

Assignment 11

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 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.