# 2002 STEP 3 Q6 (spoilers)

Please can someone check if this is the correct method:

I square rooted the expression twice - getting four equations in all. Only one of the initial conditions satisfies each one to ensure the expression under square root is real hence the eqns I obtained are:

$\sqrt(y^2 -1)$ = $\pm x+1/2$

$\sqrt(1-y^2)$ = $\pm x+1/2$

My question is the following:
For the first set of equations aren't the graphs 'half' hyperbolae since we only consider positive square root therefore domain for each graph is x>-1/2 and x<1/2 respectively

For the second set of equations aren't the graphs semi circles for same reasons as above?

Surely squaring the above equations to remove square roots would lead to extra solutions so we cannot have full hyperbolae and circles?

### Let's approach the problem a

Let's approach the problem a bit more generally. You have this equation that has powers, and so you take roots. Taking square roots can lose you solutions (for instance $x^2=9$ would lead you to $x=3$. Then you want to raise your solutions to powers, which as you correctly say will give you extra solutions.

There are a few ways round this. Sometimes the easiest is to get all the answers that could be right (i.e. square your last equation) and then check if they work. I think here they do, so you get full curves.

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