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Plus or minus on roots questions

Okay so I just wanted to confirm something

So when you have the square root of a number you get plus or minus in your answer

9=±3\sqrt{9}=\pm3

or

50=±52\sqrt{50}=\pm 5\sqrt{2}

But, its not just for square root, whenever the root is a even number you get a plus or minus ... right?

164=±2\sqrt[4]{16}=\pm2

However, if it is to the root of an odd number, you DONT use plus or minus in your answer??

273=3\sqrt[3]{27}=3

1287=2\sqrt[7]{128}=2


So would this mean that
xn\sqrt[n]{x}

For n even, give plus or minus?
For n odd, gives positive?

Is this correct? If so, how would this work for when n is negative, or a decimal, or a fraction? What is the general rule? Thanks
(edited 8 years ago)
Reply 1
Avatar for TeeEm
TeeEm
19
Original post by GarlicBread01
Okay so I just wanted to confirm something

So when you have the square root of a number you get plus or minus in your answer

9=±3\sqrt{9}=\pm3

or

50=±52\sqrt{50}=\pm 5\sqrt{2}

But, it not just for square root, whenever the root is a even number you get a plus or minus ... right?

164=±2\sqrt[4]{16}=\pm2

However, if it is to the root of an odd number, you DONT use plus or minus in your answer??

273=3\sqrt[3]{27}=3

1287=2\sqrt[7]{128}=2


So would this mean that
xn\sqrt[n]{x}

For n even, give plus or minus?
For n odd, gives positive?

Is this correct? If so, how would this work for when n is negative, or a decimal, or a fraction? What is the general rule? etc


No
read this also
The reason I've been having trouble with this is that I have been having inconsistency with my answers, for example sometimes the mark scheme has only included the positive answer and sometimes it includes the positive and negative answer and I was wondering if this was the reason. For example

Solve 1−x−2=0 1-x^{-2} = 0

x−2=1x^{-2}= 1
x=1−2x = \sqrt[-2]{1}
x=1x = 1 or x=−1 x=-1

Mark scheme gives correct answers.

However Solve 54x−3=254x^{-3} = 2
x−3=254x^{-3} = \frac{2}{54}
x=254−3x = \sqrt[-3]{\frac{2}{54}}
x=3x = 3 or x=−3 x = -3

Mark scheme says its wrong, only including the positive 3
(edited 8 years ago)
Reply 3
Avatar for Notnek
Notnek
21
Original post by GarlicBread01
Tthe reason I've been having trouble with this is that I have been having inconsistency with my answers, for example sometimes the mark scheme has only included the positive answer and sometimes it includes the positive and negative answer and I was wondering if this was the reason. For example

Solve 1−x−2=0 1-x^{-2} = 0

x−2=1x^{-2}= 1
x=1−2x = \sqrt[-2]{1}
x=1orx=−1x = 1 or x=-1

Mark scheme gives correct answers.

However Solve 54x−3=254x^{-3} = 2
x−3=254x^{-3} = \frac{2}{54}
x=254−3x = \sqrt[-3]{\frac{2}{54}}
x=3orx=−3x = 3 or x = -3

Mark scheme says its wrong, only including the positive 3

x−2=1x^{-2}=1

1x2=1\frac{1}{x^2}=1

x2=1x^2=1

There are two solutions for this : 1 and -1.


54x−3=254x^{-3} = 2

x−3=254x^{-3} = \frac{2}{54}

1x3=254\frac{1}{x^3} = \frac{2}{54}

x3=542=27x^3 = \frac{54}{2} = 27

This equation has only one solution : 3. If you cube a negative number you get a negative number so this equation only has a single positive solution.
Original post by GarlicBread01
Okay so I just wanted to confirm something

So when you have the square root of a number you get plus or minus in your answer

9=±3\sqrt{9}=\pm3

or

50=±52\sqrt{50}=\pm 5\sqrt{2}


You are rather confused - there is a sticky in this forum that you should read. But in brief:

9\sqrt{9} says "please write down the +ve number whose square is 9" - so that's +3, since (+3) x (+3) = 9

−9-\sqrt{9} says "please write down the -ve number whose square is 9" - so that's -3, since (-3) x (-3) = 9

x2=9x^2=9 says "please find all of the numbers whose square is 9" - as we have seen, there are two of those, so we write x=+3=9x=+3=\sqrt{9} or x=−3=−9x=-3=-\sqrt{9}, or we join those two together and write x=±9x=\pm\sqrt{9}

However, we *never* write 9=−3\sqrt{9}=-3 or 9=±3\sqrt{9}=\pm3 since the symbol
Unparseable latex formula:

\sqrt

is an order to write down a single, *positive* number.
Reply 5
Avatar for Notnek
Notnek
21
Original post by GarlicBread01
The reason I've been having trouble with this is that I have been having inconsistency with my answers, for example sometimes the mark scheme has only included the positive answer and sometimes it includes the positive and negative answer and I was wondering if this was the reason. For example

Solve 1−x−2=0 1-x^{-2} = 0

x−2=1x^{-2}= 1
x=1−2x = \sqrt[-2]{1}
x=1x = 1 or x=−1 x=-1

Mark scheme gives correct answers.

However Solve 54x−3=254x^{-3} = 2
x−3=254x^{-3} = \frac{2}{54}
x=254−3x = \sqrt[-3]{\frac{2}{54}}
x=3x = 3 or x=−3 x = -3

Mark scheme says its wrong, only including the positive 3

In addition to my previous post:

The equation x2=1x^2=1 has two solutions : 1 and -1.

But 1=1\sqrt{1}=1 only.

1=±1\sqrt{1}=\pm 1 is incorrect.

Please read this.


Hopefully this doesn't confuse you more :smile:
Original post by notnek
x−2=1x^{-2}=1

1x2=1\frac{1}{x^2}=1

x2=1x^2=1

There are two solutions for this : 1 and -1.


54x−3=254x^{-3} = 2

x−3=254x^{-3} = \frac{2}{54}

1x3=254\frac{1}{x^3} = \frac{2}{54}

x3=542=27x^3 = \frac{54}{2} = 27

This equation has only one solution : 3. If you cube a negative number you get a negative number so this equation only has a single positive solution.


Okay so does this mean that when it is even, (e.g square root, 4th root etc.), there are two solutions, 'plus and minus' and when it is an odd, (e.g cube root, 5th root etc.) then there is one solution, the positive solution?

@Edit: I just saw your replys and I will read the sticky you linked.
(edited 8 years ago)
Original post by GarlicBread01
Okay so does this mean that when it is even, (e.g square root, 4th root etc.), there are two solutions, 'plus and minus' and when it is an odd, (e.g cube root, 5th root etc.) then there is one solution, the positive solution?


http://www.mathopenref.com/rootnumber.html
Original post by Kvothe the arcane


Thank you so much! The website really helped clarify the confusion I was having.

Although the website explained what to do for odd and even degrees, and I assume the same applies for negative degrees. The only other confusion that I can think of left, is how would the rule work for decimal/fractions seeing as they are neither odd or even? Do you just take the positive answer?

For example

x0.5\sqrt[0.5]{x}

or

x23\sqrt[\frac{2}{3}]{x}
Original post by GarlicBread01
Thank you so much! The website really helped clarify the confusion I was having.

Although the website explained what to do for odd and even degrees, and I assume the same applies for negative degrees. The only other confusion that I can think of left, is how would the rule work for decimal/fractions seeing as they are neither odd or even? Do you just take the positive answer?

For example

x0.5\sqrt[0.5]{x}

or

x23\sqrt[\frac{2}{3}]{x}

Despite what the link says (and I'm glad it gives you some clarity), you must understand that when you write down something to the power of something else or something root something, you're referring to a specific number. It's another thing to find the roof of an equation.

It would depend on the fraction. In the case negative indices, that refers to the variable in the denominator so you'd manipulate the equation till it became the subject.

If you raise the power of the equation to the denominator of a fractional index of a variable then you're left with x to the power of something is = to something else. Whether the number is even or odd will make you aware of how many solutions there are.

x3/2=4⇒x3=42=16⇒x=163x^{3/2}=4 \Rightarrow x^3=4^2=16 \Rightarrow x=\sqrt[3]{16}
x2/3=4⇒x2=64⇒x=±8x^{2/3}=4 \Rightarrow x^2=64 \Rightarrow x= \pm 8
(edited 8 years ago)
I see, I'm just so used to hearing stuff like "the square root of a number is plus or minus" that I've gotten confused but instead only applies for say

when solving x2=9x^2 = 9
it is correct to sayx=9x = \sqrt 9
x=±3x = \pm 3

but it would be incorrect to say

9=±3\sqrt 9 = \pm 3 from "nowhere" without an equation
Original post by GarlicBread01
I see, I'm just so used to hearing stuff like "the square root of a number is plus or minus" that I've gotten confused but instead only applies for say

when solving x2=9x^2 = 9
it is correct to sayx=9x = \sqrt 9
x=±3x = \pm 3

but it would be incorrect to say

9=±3\sqrt 9 = \pm 3 from "nowhere" without an equation


The square root of a number refers to the positive root.

9≠±3\sqrt{9} \neq \pm 3 however you are correct in saying that there are two solutions to the equation x2=9x^2=9, namely x=3 or−3x = 3 \ \text{or} -3. This is by convention though. I've heard it said differently.
(edited 8 years ago)
Reply 12
Avatar for Notnek
Notnek
21
Original post by GarlicBread01
I see, I'm just so used to hearing stuff like "the square root of a number is plus or minus" that I've gotten confused but instead only applies for say

when solving x2=9x^2 = 9
it is correct to sayx=9x = \sqrt 9
x=±3x = \pm 3

but it would be incorrect to say

9=±3\sqrt 9 = \pm 3 from "nowhere" without an equation

Nearly but it should be

x2=9x^2 = 9
x=±9x = \pm\sqrt 9

Unparseable latex formula:

\sqrt

always refers to the positive square root only.


It is extremely common for students to be confused by this. I did a mini study on this area a few years back and was amazed by how many teachers and textbooks confuse students with statements like 9=±3\sqrt{9}=\pm 3. The old edition of the edexcel C1 textbook contained this exact equation until it was corrected in the later edition.
(edited 8 years ago)
Original post by notnek
Nearly but it should be

x2=9x^2 = 9
x=±9x = \pm\sqrt 9

Unparseable latex formula:

\sqrt

always refers to the positive square root only.


It is extremely common for students to be confused by this. I did a mini study on this area a few years back and was amazed by how many teachers and textbooks confuse students with statements like 9=±3\sqrt{9}=\pm 3. The old edition of the edexcel C1 textbook contained this exact equations until it was corrected in the later edition.


Certainly textbooks differ on whether they say the term square root refers to the positive root of such an equation.
Reply 14
Avatar for Notnek
Notnek
21
Original post by Kvothe the arcane
Certainly textbooks differ on whether they say the term square root refers to the positive root of such an equation.

The term "square root" is a bit more wishy-washy than the square root symbol
Unparseable latex formula:

\sqrt

which should always be the positive square root.

You could say that -3 and 3 are both "square roots" of 9. But using the word "the" e.g. "the square root of 9" usually means the positive square root.
Original post by GarlicBread01
The reason I've been having trouble with this is that I have been having inconsistency with my answers, for example sometimes the mark scheme has only included the positive answer and sometimes it includes the positive and negative answer and I was wondering if this was the reason. For example

Solve 1−x−2=0 1-x^{-2} = 0

x−2=1x^{-2}= 1
x=1−2x = \sqrt[-2]{1}
x=1x = 1 or x=−1 x=-1

Mark scheme gives correct answers.

However Solve 54x−3=254x^{-3} = 2
x−3=254x^{-3} = \frac{2}{54}
x=254−3x = \sqrt[-3]{\frac{2}{54}}
x=3x = 3 or x=−3 x = -3

Mark scheme says its wrong, only including the positive 3


X^3 is one to one.
Similarly square root function is one to one.
Quadratic isnt these.
Sqrt(x^2)=|x|={x for x>=0,-x x<0}





Posted from TSR Mobile
Reply 16
Avatar for Zacken
Zacken
22
Original post by Kvothe the arcane
Certainly textbooks differ on whether they say the term square root refers to the positive root of such an equation.


As, notnek said - a number may have 2 square roots but the square root of a number is the positive square root, this is because we define f:R+→R+f: \mathbb{R}^{+} \to \mathbb{R}^{+} for the map x↦xx \mapsto \sqrt{x}.

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