q: in the equation √(x +3) = x – 3, why isn’t x = 1 a valid solution?  after all, isn't the square root of 4 also negative 2?

a: this is a very common issue for students preparing for the gre. let's talk about it :)

generally, a square root question will have two solutions, one positive and one negative. for instance, the square root of 4 is 2 and -2.

on the gre, if the "√" symbol appears as part of the question, we can only use the positive answer of the square root.

but if the "√" symbol is not part of the question, then both the negative and positive roots of a square root can be used.

the radical sign (more commonly known as the square root sign) actually means the positive square root of a given number. this positive square root is often referred to as the principal square root. anytime we see the radical sign on the gre, we want the principal (positive) square root.

also, from wikipedia:

"every non-negative real number a has a unique non-negative square root, called the principal square root, which is denoted by [square root symbol], where [square root symbol] is called radical sign. for example, the principal square root of 9 is 3, denoted sqrt(9)=3 , because 3^2=3×3=9 and 3 is non-negative."

also check out this blog on square roots. we go into detail regarding when you do and do not consider the negative values in questions like these.

q: for the first practice problem, how come having a square root of a negative number on each side of the equation results in no solution?

a: in the lesson video square roots, mike talks about trying to take the square root of a negative number.  he specifically says the following:

can we take the square root of a negative? no. nothing on the number line can be squared to yield a negative number. now, as it turns out, there are higher forms of mathematics where they talk about square roots of a negative number, and that's called imaginary numbers. 

you do not need to worry about this for the test. that is absolutely beyond anything that is on the test. 

and according to ets’s mathematical conventions document, “all numbers used in the test questions are real numbers. in particular, integers and both rational and irrational numbers are to be considered, but imaginary numbers are not.”

thus, for the first practice problem, we end up with no solution because when we plug in the value of -2 for x, we end up with imaginary numbers on both sides of the equation.  and, on the gre, we can’t do math with that.  thus, for our purposes, that is just an error, and this equation has no solution.

q: for the second practice problem, how did we get x= 0 and x=1 from 0 = x^2 - x?

a: our first step is to simplify the expression above to:

0 = x(x-1)

in order to solve for the different values of x from here, we can consider the two expressions of x separately. in other words:

• 0 = x
• 0 = x-1

to see why we can do this, it's key to keep in mind that any number multiplied by 0 is equal to 0. yes, that's a pretty basic fact, but applied in equations like this, it is important to keep in mind!

first, let's plug in x = 0:

• x(x-1) --> 0(0-1) = 0*(-1) = 0

as we can see, x=0 satisfies the original equation, which says that the product of the two expression equals 0. likewise, when x-1 = 0, we can write,

• x(x-1) --> x*0 = 0.

therefore, x-1 = 0 also satisfies the original equation. solving for x, we get x=1.

based on this method, our two solutions are: x=0 and x=1.

q: should we always check for extraneous roots whenever we solve an algebraic equation?

a: good question! there are only two times you'll have to worry about extraneous roots on the gre:

1. when you have a variable under a radical
2. when you have a variable within an absolute value

in quadratic equations without radicals, you don't need to worry about extraneous roots :)