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Schachfreizeit
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Square root 26 Dec 2022, 22:56
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If in the gmat the square root of an integer is always positive, then why is √x=|x| and not just x?

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Square root 27 Dec 2022, 01:07
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Schachfreizeit
If in the gmat the square root of an integer is always positive, then why is √x=|x| and not just x?
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I think you meant to write \(\sqrt{x^2}=|x|\), not \(√x=|x|\).

About \(\sqrt{x^2}=|x|\). Notice that the square root there gives, as it should, non-negative result: |x| (the absolute value of a number is always nonnegative). Consider this, say we have \(\sqrt{x^2}=5\). What is the value of x? Well, x can obviously be 5: \(\sqrt{5^2}=\sqrt{25}=5\) but it can also be -5: \(\sqrt{(-5)^2}=\sqrt{25}=5\). So, as you can see \(\sqrt{x^2}=5\) means that \(\sqrt{x^2}=|x|=5\), which gives x = 5 or x = -5.

Hope it's clear.
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Square root 27 Dec 2022, 08:50
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Schachfreizeit
If in the gmat the square root of an integer is always positive, then why is √x=|x| and not just x?

I think you meant to write \(\sqrt{x^2}=|x|\), not \(√x=|x|\).

About \(\sqrt{x^2}=|x|\). Notice that the square root there gives, as it should, non-negative result: |x| (the absolute value of a number is always nonnegative). Consider this, say we have \(\sqrt{x^2}=5\). What is the value of x? Well, x can obviously be 5: \(\sqrt{5^2}=\sqrt{25}=5\) but it can also be -5: \(\sqrt{(-5)^2}=\sqrt{25}=5\). So, as you can see \(\sqrt{x^2}=5\) means that \(\sqrt{x^2}=|x|=5\), which gives x = 5 or x = -5.

Hope it's clear.
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Thank you!
My problem is that I thought if we take the square root of a certain value in the gmat, only the positive solution counts and we neglect the negative solution
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Square root 27 Dec 2022, 08:55
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Schachfreizeit
Bunuel
Schachfreizeit
If in the gmat the square root of an integer is always positive, then why is √x=|x| and not just x?

I think you meant to write \(\sqrt{x^2}=|x|\), not \(√x=|x|\).

About \(\sqrt{x^2}=|x|\). Notice that the square root there gives, as it should, non-negative result: |x| (the absolute value of a number is always nonnegative). Consider this, say we have \(\sqrt{x^2}=5\). What is the value of x? Well, x can obviously be 5: \(\sqrt{5^2}=\sqrt{25}=5\) but it can also be -5: \(\sqrt{(-5)^2}=\sqrt{25}=5\). So, as you can see \(\sqrt{x^2}=5\) means that \(\sqrt{x^2}=|x|=5\), which gives x = 5 or x = -5.

Hope it's clear.


Thank you!
My problem is that I thought if we take the square root of a certain value in the gmat, only the positive solution counts and we neglect the negative solution
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Yes, that's true: when we have the square root sign for an even root, such as a square root, fourth root, etc. then the only accepted answer is the non-negative root. That is:

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

That is not violated with \(\sqrt{x^2}=|x|\) because |x| is the absolute value of x, which is always nonnegative.
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Square root 27 Dec 2022, 10:04
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Hi Schachfreizeit

Iam sure you meant what has already been pointed out by Bunuel

Let me try to put my 0.02$ here! :think:

:idea: What is the degree of \(x^2 \)=49?
Degree is the highest power of the variable in an equation.
The cardinal relationship between degree and roots(solutions or values that satisfy the equation) of an equation is the following-
Degree of an equation = Number of roots of the equation.

So, how many roots must \(x^2 \)=49 have? - Its 2 Right?
What all values can you think that satisfies \(x^2 =49\) ? 2 values+ 7 and -7 Right?
So the first point I would like to bring to your notice is

\(\sqrt{(variable)^2}\) = +/- variable

Think about it this way- If \(\sqrt{x^2} \)is the expression then, what can I have inside the square root sign to have an outcome that can be 0 or positive?

:idea: But what if I ask you to find \(\sqrt{49} \)?
Let x= \(\sqrt{49}\)
What is the degree of this equation? Its 1=> Linear equation
It has to have only 1 root.
So the second point I would like to bring to your notice is -

\( \sqrt{constant }\) = positive value/0

Even if I try to plot the graphs for \(x^2\) =49 and x=\( \sqrt{49},\) you will have a curve for the first function with degree 2 and a straight line for the second equation with degree 1.
A curve is NOT equal to a straight line.

:idea: Mod |x| = +x when x>0 and -x when x<0. Its also =0 when x=0.
So |x| =+/-x
Can you connect the dots now?
|x|= \(\sqrt{x^2}\) since \(\sqrt{x^2}\) = +/- x
Hope this difference is clear to you. :)

:idea: PS-Its an extremely important concept and a test taker must be super clear on this.

Devmitra Sen
GMAT Mentor

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