Surds Question

SignTheContract

Hi there. I was just looking a t a past question in the LC and don't really understand the last bit. Here's the question:

Solve sqrt(2x + 5) - 1 = x

So squaring gives 2x+5 = x^2 + 2x + 1

leaves us with x^2 - 4 = 0

solving gives x = 2 or x = -2

I'm fine with all the above but the book tells me to sub both numbers back into the original question to check.

so for x = 2 we have sqrt(2(2)+5) = 2 + 1
=sqrt(9)=3
=3=3 so thats fine.

But for x = -2 we have sqrt(2(-2)+5)=-2+1
=> sqrt(1) = -1

The book now says that this is not a solution as sqrt(1) does not = -1. But I thought every postive number had two square roots. So the sqrt of 1 is also -1 so what am I not getting?

Fremen

It's true that for any positive real number x, there exist two real numbers, y and -y, such that y^2 = (-y)^2 = x.
However, by convention, the sqrt(.) function means the positive solution of the equation above.

If you think about it, it's totally necessary to make the distinction. Otherwise Sqrt(x) and -Sqrt(x) would mean the same thing, or would be ambiguous at best.

bnt

As you say, sqrt(1) = ±1 . If you square both sides of sqrt(1) = -1, you get 1 = 1, so I don't see a problem there. It just looks weird.

CJC86

I'd have to disagree bnt. If we are talking about the function sqrt(.), then it is important that it is single-valued.

Otherwise, you get into all sorts of trouble when you start to generalise x^y for a real number y. Any rational number y=m/n would have n values, and we couldn't really have functions such as x^(pi) since we would not have any sensible convergence. (This is just expanding on Fremen's point tbh)

For this reason it is important that functions are single-valued, so for sqrt(x) we choose the "principal value", which is the positive real number that when squared gives us x. It may seem a bit pointless at Leaving Cert level, but there is a very sensible reason for making the distinction between sqrt(x) and a number y such that y^2=x.

Michael Collins

As Freman has said the square root function denoted by the radical sign i.e.

[latex] \displaystyle f(x) = \sqrt(x) [/latex]

is single valued (otherwise it wouldn't be a function), and always equals the positive square root when talking about roots of positive real numbers.

So

[latex] \displaystyle \sqrt(1) = 1 [/latex]

or more generally for any real number x

[latex] \displaystyle \sqrt(x^2) = |x| [/latex]

Michael Collins

Ha, you just beat me to it CJC86. And you covered all my points, but better!

SignTheContract

Thanks a million folks. That makes a bit more sense now. Really appreciated the help! I'll probably be back on this forum for more help during the year!

bnt

Well, I'm going to argue with that level of abstraction. I can also see what you mean if I think terms of complex maths, in a 2D phasor representation, where a resultant phasor always has a positive magnitude. A negative magnitude would be nonsensical in that context.