Confusion on Square Roots : GMAT Quantitative Section
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# Confusion on Square Roots

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Manager
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19 Feb 2012, 12:22
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.
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Re: Confusion on Square Roots [#permalink]

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19 Feb 2012, 12:26
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Expert's post
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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Re: Confusion on Square Roots [#permalink]

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19 Feb 2012, 14:55
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I think the difference is in variable and constant

[square_root][25] will always be 5

but sq rt [x2] = abs [x] can be +/- x depending on whether x is positive or negative, x being a variable.

Do you think the above reasoning is correct?
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Re: Confusion on Square Roots [#permalink]

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19 Feb 2012, 21:45
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raviram80 wrote:
I think the difference is in variable and constant

[square_root][25] will always be 5

but sq rt [x2] = abs [x] can be +/- x depending on whether x is positive or negative, x being a variable.

Do you think the above reasoning is correct?

No that's not correct. It seems that you should brush up your fundamentals. Please check Absolute value and Number Theory chapters of Math Book for that: math-absolute-value-modulus-86462.html and math-number-theory-88376.html

Now, $$\sqrt{x^2}=|x|$$, but absolute value is ALWAYS nonnegative (since it basically measures the distance and distance cannot be negative), so $$\sqrt{x^2}=|x|=nonnegative$$ as it should be.

As for $$|x|$$: if $$x<0$$ (so when $$x$$ is negative) then $$|x|=-x=-negative=positive$$ and if $$x>0$$ (so when $$x$$ is positive) then $$|x|=x=positive$$. As you can see $$|x|$$ is positive in both cases ($$|x|=0$$ when $$x=0$$).

Hope it helps.
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Re: Confusion on Square Roots [#permalink]

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22 Feb 2012, 08:20
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raviram80 wrote:
[square_root][25] will always be 5

This is where your line of reasoning fails.

$$sqrt(25)$$ yields a number which when multiplied by itself equals $$25$$.
In this case, there are two numbers which when multiplied by themselves equal 25:

- $$+5$$ when multiplied by itself equals $$25$$
- $$-5$$ when multiplied by itself equals $$25$$

As a result, $$sqrt(25)$$ has two solutions: $$5$$ and $$-5$$.

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Re: Confusion on Square Roots [#permalink]

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22 Feb 2012, 08:26
fxsunny wrote:
raviram80 wrote:
[square_root][25] will always be 5

This is where your line of reasoning fails.

$$sqrt(25)$$ yields a number which when multiplied by itself equals $$25$$.
In this case, there are two numbers which when multiplied by themselves equal 25:

- $$+5$$ when multiplied by itself equals $$25$$
- $$-5$$ when multiplied by itself equals $$25$$

As a result, $$sqrt(25)$$ has two solutions: $$5$$ and $$-5$$.

$$\sqrt{25}=5$$, NOT +5 or -5. Please refer to my first post.
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Re: Confusion on Square Roots   [#permalink] 22 Feb 2012, 08:26
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