Square Root = Always Positive?

\(x^2 = positive \ number\) ALWAYS has TWO solutions: \(x = \sqrt{positive \ number}\) and \(x = -\sqrt{positive \ number}\).

For example, x^2 = 9 --> \(x = \sqrt{9}=3\) or \(x = -\sqrt{9}=-3\).

\(p^2=(q+1)^2\) means that \(|p|=|q+1|\), so p=q+1 or p=-(q+1)

P.S. What DS question are you talking about?

Q: If p and q are positive integers, is the greatest common factor of p and q greater than 1?
1: \(p^2 = pq + p\)
2: \(p^2 = q^2 + 2q + 1\)

So you mean to say, if GMAT gives you square root, we take only positive.
But when it gives square like \(x^2 = 9\) , we take both values?

Notice that in the question we are told that p and q are POSITIVE. So, here from \(p^2=(q+1)^2\) it follows that \(p=q+1\) only.

Finally, to answer your question, check the discussion on page 1. For example, this one: https://gmatclub.com/forum/square-root- ... l#p1437877

When the GMAT provides the square root sign for an even root, such as a square root, fourth root, etc. then the only accepted answer is the positive root. That is:

\(\sqrt{9} = 3\), NOT +3 or -3;
\(\sqrt[4]{16} = 2\), NOT +2 or -2;

Notice that in contrast, the equation \(x^2 = 9\) has TWO solutions, +3 and -3. Because \(x^2 = 9\) means that \(x =-\sqrt{9}=-3\) or \(x=\sqrt{9}=3\).

Hi VeritasKarishma

i wonder why arent you considering two cases for this inequality \(\sqrt{(x+1)^2} <= \sqrt{36}\)

|x+1| <= 6

i thought there should be TWO cases to consider - positive and negative x+1<= 6 and x+1 >= -6
because \(\sqrt{x^2} =|x|\) which means x can be negative or positive

Just like with this equation (x-1)^2=400

Another question what if we had such inequity (10-x)^2 <= 9 would there be two cases ? i lay stress in this inequality on X being negative, very doubt if inequality is correct

thanks

Those are two different things everywhere, not just on the GMAT. I bring this up because there seems to be a persistent misconception in GMAT prep circles that there is such a thing as "GMAT math" -- a set of math rules or conventions you need to learn specifically for this test. That's not true; GMAT math is exactly the same as the math you'd learn in any other math course (within the bounds the GMAT specifies, e.g. that all numbers are real numbers), so everything test takers learn about math during GMAT prep will potentially also be useful when they're doing math in their MBA or elsewhere.


It looks like Karishma didn't finish solving the problem in that post. If you do see the inequality:

|x + 1| < 6

then x+1 must be somewhere between -6 and 6 (inclusive), so x itself is somewhere between -7 and 5 inclusive. It's certainly possible that x is negative, and you must consider that possibility to get the right solution set.

Similarly if you know (10 - x)^2 < 9, then 10 - x must be somewhere between -3 and 3 inclusive. If you then simplify the two inequalities 10 - x > -3 and 10 - x < 3, you find 7 < x < 13.

The radical sign \(\sqrt{x}\), when applied to a real number, denotes the principal or positive square root.

=> If x ≥ 0 then \(\sqrt{x}\) ≥ 0.

The square root of 'x' is defined as a number whose square is 'x' and hence that may be positive or negative according to algebra; whereas, sqrt termed as 'principal square root' defines that a positive number whose square is 'x.


Let's analyze x = \(\sqrt{16}\)

Plug x = 4: 4 = \(\sqrt{16}\) - It is TRUE

Plug x = -4: -4 = \(\sqrt{16}\) - It is FALSE


In \(x^2\) = 16

Plug x = 4: \(4^2\) = 16 - It is TRUE

Plug x = -4: \((-4)^2\) = 16 - It is TRUE

Thanks

I explain it at 4:45 in this video:
Subscribe for more: https://www.youtube.com/QuantReasoning? ... irmation=1

Sorry. Late to the party but just spotted this as it came up in class this week. There are always 2 square roots to a positive number; one positive, one negative. The squiggle that goes right over the top of the number means 'Square root' (2 values). The shorter symbol which just goes down and up but does not go over the top of the number indicates the +ve square root only.

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but if the square root has only one solution, then √x'2 must be x and not |x|...
where is my error in thinking?

take the example of x = -4

x^2 = 16, so sqrt(16) = 4 , which is -x

thats why sqrt(x^2) = |x|