raviram80 wrote:

Hi All,

MGMAT guide says that sq rt (x^2) = abs val (x)

e.g x^2 =25 -> x = +/- 5

abs (5) = +/- 5

Now when we solve this eqn

sq rt (x) = x-2 we get two solns x=4 and x=1

The way it shows to confirm is by putting these back in the eqn, so

sq rt (4) = 4-2

2 =2 , so correct

sq rt (1) = 1-2

1 = -1 so incorrect and so only 4 is the solution

My question

why sq rt (4) not equal to sq rt (2^2) and as from above +/-2 is the solution

and sq rt (1) not equal, to sq rt (1^1) and as from above +/-1 is the solution

In the above cases I mention sq rt (1^1) = -1 satisfies the eqn.

I am confused on what is the process and logic here to solve GMAT even roots problems.

SOME NOTES:1. GMAT is dealing only with

Real Numbers: Integers, Fractions and Irrational Numbers.

2. Any nonnegative real number has a

unique non-negative square root called

the principal square root and unless otherwise specified,

the square root is generally taken to mean

the principal square root.

When the GMAT provides the square root sign for an even root, such as \(\sqrt{x}\) or \(\sqrt[4]{x}\), then the

only accepted answer is the positive root.

That is, \(\sqrt{25}=5\), NOT +5 or -5. In contrast, the equation \(x^2=25\) has TWO solutions, \(\sqrt{25}=+5\) and \(-\sqrt{25}=-5\).

Even roots have only non-negative value on the GMAT.Odd roots will have the same sign as the base of the root. For example, \(\sqrt[3]{125} =5\) and \(\sqrt[3]{-64} =-4\).

As for \(\sqrt{x}\) and \(x^{\frac{1}{2}}\): they are the same.

3. \(\sqrt{x^2}=|x|\).

The point here is that as

square root function can not give negative result then \(\sqrt{some \ expression}\geq{0}\).

So \(\sqrt{x^2}\geq{0}\). But what does \(\sqrt{x^2}\) equal to?

Let's consider following examples:

If \(x=5\) --> \(\sqrt{x^2}=\sqrt{25}=5=x=positive\);

If \(x=-5\) --> \(\sqrt{x^2}=\sqrt{25}=5=-x=positive\).

So we got that:

\(\sqrt{x^2}=x\), if \(x\geq{0}\);

\(\sqrt{x^2}=-x\), if \(x<0\).

What function does exactly the same thing? The absolute value function: \(|x|=x\), if \(x\geq{0}\) and \(|x|=-x\), if \(x<0\). That is why \(\sqrt{x^2}=|x|\).

Hope it's clear.

_________________

NEW TO MATH FORUM? PLEASE READ THIS: ALL YOU NEED FOR QUANT!!!

PLEASE READ AND FOLLOW: 11 Rules for Posting!!!

RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory; 7. Remainders; 8. Overlapping Sets; 9. PDF of Math Book; 10. Remainders; 11. GMAT Prep Software Analysis NEW!!!; 12. SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) NEW!!!; 12. Tricky questions from previous years. NEW!!!;

COLLECTION OF QUESTIONS:

PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.

What are GMAT Club Tests?

25 extra-hard Quant Tests

GMAT Club Premium Membership - big benefits and savings