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.

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