Does the square root of 9 equal 3? Or can it also equal negative 3? Ask these questions to a GCSE exam ready student and most should answer with the latter. However, caution should be exercised when discussing this point as confusion can arise depending on the context and notation used.
The issue arises with the use of the radical symbol (√) as a way of stating the square root of a number. But what is the difference? As a concept, the square root of 9 is a number that when squared equals 9. Clearly, two numbers fit the criteria, those being 3 and -3. The use of the radical symbol is used to represent a specific value. √9=3 is a correct statement whereas √9=±3 is not, unless used as a line of working solutions to a quadratic equation, which could be the crux of where the confusion lies.
Asking to find the solutions to x^2=9 and to find the value of √9 are separate things. Think about this when working with surds. Consider the following:
√5+√5
When simplifying this surd expression you wouldn’t consider each term to have two separate values. Such recklessness would yield three separate values, i.e. -√5,0 and √5. Maybe the fact that each term is irrational stops you entertaining the idea of it having a positive and negative value. There is notation that can be used for this by inserting a ± in front of the radical symbol, most famously used as part of the Quadratic Equation.
So why does this matter? Well, it can cost marks in an exam, maybe not so much at GCSE, but certainly at A-Level. For example, if you need to evaluate the value of a function which is composed of a root, then you certainly need to make sure your answer does not consist of multiple values. Another thing to think about is when you need to square both sides of an equation to solve. Consider the equation below:
3-√(x+5)=7
Adding the radical to both sides, subtracting 7 and then squaring gets you to 16=x+5 which follows to a solution of x=11. You should note, however, (and this will probably need a separate blog post on it’s own) that x=11 does not satisfy the equation, unless you wrongly assume that the √(x+5) has a positive and negative value, and so the equation has no real solutions.