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# Confusion on Square Roots

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Manager
Joined: 09 Jul 2010
Posts: 74

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19 Feb 2012, 13:22
2
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.
Math Expert
Joined: 02 Sep 2009
Posts: 57996
Re: Confusion on Square Roots  [#permalink]

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19 Feb 2012, 13:26
5
3
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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Manager
Joined: 09 Jul 2010
Posts: 74
Re: Confusion on Square Roots  [#permalink]

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19 Feb 2012, 15:55
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?
Math Expert
Joined: 02 Sep 2009
Posts: 57996
Re: Confusion on Square Roots  [#permalink]

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19 Feb 2012, 22:45
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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Joined: 13 Jan 2012
Posts: 36
Re: Confusion on Square Roots  [#permalink]

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22 Feb 2012, 09:20
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 $$-5$$.

Math Expert
Joined: 02 Sep 2009
Posts: 57996
Re: Confusion on Square Roots  [#permalink]

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22 Feb 2012, 09:26
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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Manager
Joined: 09 Aug 2016
Posts: 63
Re: Confusion on Square Roots  [#permalink]

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27 Jan 2017, 05:43
Is any topic with lets say 10 examples covering all corner cases for the sake of completeness?
Manager
Joined: 16 May 2018
Posts: 85
Location: Hungary
Schools: Queen's MBA'20
Re: Confusion on Square Roots  [#permalink]

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02 Aug 2019, 12:55
I have a question pls:
in case: √36 = x
Does it mean that x= +- 6 or only +6 ?
Math Expert
Joined: 02 Sep 2009
Posts: 57996
Re: Confusion on Square Roots  [#permalink]

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02 Aug 2019, 14:26
hsn81960 wrote:
I have a question pls:
in case: √36 = x
Does it mean that x= +- 6 or only +6 ?

√36 = 6 only. The square root cannot give negative result. Check above posts for more.
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Re: Confusion on Square Roots   [#permalink] 02 Aug 2019, 14:26
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