Is it Possible for x^2 to Exceed 900?

In summary: The best (and only) way to solve this problem is to consider two cases and see where each case takes you.
  • #1
mark2142
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Homework Statement
Suppose ##x^2 > 900##
Relevant Equations
What will be x?
We can also write a general inequality ##x^2>a## where a is a number.
If ##x^2= a##, then ##x= \pm \sqrt a## which means ##x= \sqrt a## or ##-\sqrt a##.
But in this case i don’t think it will be ##x > \pm \sqrt a## because if we take ##0## , it’s greater than a negative but in the original inequality ##0 < a## , a positive number.
 
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  • #2
How about ##|x|## ?
 
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  • #3
mark2142 said:
If ##x^2= a##, then ##x= \pm \sqrt a## which means ##x= \sqrt a## or ##-\sqrt a##.
But in this case i don’t think it will be ##x > \pm \sqrt a## ...
That's true. ##x^2 > a \Rightarrow x > a## or ##x \ ? -a##?
 
  • #4
mark2142 said:
Homework Statement:: Suppose ##x^2 > 900##
Relevant Equations:: What will be x?

We can also write a general inequality ##x^2>a## where a is a number.
If ##x^2= a##, then ##x= \pm \sqrt a## which means ##x= \sqrt a## or ##-\sqrt a##.
But in this case i don’t think it will be ##x > \pm \sqrt a## because if we take ##0## , it’s greater than a negative but in the original inequality ##0 < a## , a positive number.
##x > \pm \sqrt a## is ambiguous and shouldn’t be used.

For example, if ##a=900## then one possible interpretation of ##x > \pm \sqrt a## is that it is satisfied by ##x=1##, because ##1>-30##, which is clearly wrong does not satisfy the original inequality##x^2 \gt a##.

Edited.
 
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  • #5
mark2142 said:
But in this case i don’t think it will be ##x > \pm \sqrt a## because if we take ##0## , it’s greater than a negative but in the original inequality ##0 < a## , a positive number.
Good catch. When you are dealing with inequalities, you have to be careful about minus signs and negative numbers. You should handle these two cases separately.
Case 1: ##x \gt \sqrt a##
Case 2: ##x \lt -\sqrt a##
 
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  • #6
Dear @mark2142 ,

Are you aware that if ## a > b ##, then it follows that ## - a < - b## ?

I.e. if you multiply left and right of an inequality with a negative number : the inequality sign flips ...

##\ ##
 
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  • #7
Steve4Physics said:
##x > \pm \sqrt a## is ambiguous and shouldn’t be used.

For example, if ##a=900## then one possible interpretation of ##x > \pm \sqrt a## is that it is satisfied by ##x=1##, because ##1>-30##, which is clearly wrong does not satisfy the original inequality##x^2 \gt a##.

Edited.
Yes! ##x> - \sqrt a## is wrong and then ##x^2 > a## is not true.
But how do we reach to the solution?
(I don’t want to solve ##x^2 -a > 0##.)
 
  • #8
FactChecker said:
Good catch. When you are dealing with inequalities, you have to be careful about minus signs and negative numbers. You should handle these two cases separately.
Case 1: ##x \gt \sqrt a##
Case 2: ##x \lt -\sqrt a##
How have you reached to the solution?
( sorry for late response! )
 
  • #9
BvU said:
How about ##|x|## ?
What about it?
 
  • #10
mark2142 said:
Yes! If ##x> - \sqrt a## then ##x^2 > a## is not true.
But how do we reach to the solution?
(I don’t want to solve ##x^2 -a > 0##.)
Come again ? In post #1 you ask ' what is ##x## if ##x^2>900## ' and now you don't want to solve the equivalent problem ?

How about making a plot, then ?

1679052738213.png

##\ ##
 
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  • #11
mark2142 said:
What about it?
##x^2 > 900\ \Leftrightarrow \ |x|>\sqrt{900}##
 
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  • #12
BvU said:
Come again ? In post #1 you ask ' what is ##x## if ##x^2>900## ' and now you don't want to solve the equivalent problem ?

How about making a plot, then ?

View attachment 323730
##\ ##
I thought maybe there was a fast algebraic method. So tell me about the graph.You plotted a graph between ##x## and ##x^2## by taking different values of x. By graph it’s clear ##x^2> 900## for ##x <-30## and ##x>30##.
 
  • #13
mark2142 said:
I thought maybe there was a fast algebraic method
There is:
BvU said:
##x^2 > 900\ \Leftrightarrow \ |x|>\sqrt{900}##
##\Leftrightarrow \ x>30\ \lor \ -x<-30 ##

[edit] Oops ! see #16
 
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  • #14
mark2142 said:
I thought maybe there was a fast algebraic method.
Not really. The best (only?) way is to consider two cases and see where each case takes you.
 
  • #15
FactChecker said:
Not really. The best (only?) way is to consider two cases and see where each case takes you.
Ok! Thanks everyone. Bye!
 
  • #16
BvU said:
##\Leftrightarrow \ x>30\ \lor \ -x<-30 ##
##\Leftrightarrow \ x>30\ \lor \ x<-30 ##
 
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  • #17
Emphasis added...
mark2142 said:
So tell me about the graph. You plotted a graph between ##x## and ##x^2## by taking different values of x. By graph it’s clear ##x^2> 900## for ##x <-30## and ##x>30##.
  1. @BvU provided a graph of ##y = x^2## and a graph of ##y = 900## to demonstrate visually the intervals for which ##x^2 > 900##.
  2. The appropriate conjunction above is or. It's not possible for a value of x to be both less than -30 and greater than 30.
 
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  • #18
FactChecker said:
Not really. The best (only?) way is to consider two cases and see where each case takes you.
I think the OP is asking how you come up with the two cases in the first place. That is, how do you know it's supposed to be ##x < -\sqrt{a}## instead of ##x > -\sqrt{a}## other than simply memorizing?

The easiest way is using what @BvU has hinted at. Since ##\sqrt{x^2} = \lvert x \rvert##, after taking the square root of the original inequality, you have ##\lvert x \rvert > 30##. Then the definition of the absolute value leads to the two cases
$$ \lvert x \rvert > 30 \quad \Rightarrow \quad
\begin{cases}
x > 30, & x>0 \\
-x > 30, & x<0
\end{cases}$$ Negating the latter case gives you ##x < -30##.
 
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  • #19
vela said:
I think the OP is asking how you come up with the two cases in the first place. That is, how do you know it's supposed to be ##x < -\sqrt{a}## instead of ##x > -\sqrt{a}## other than simply memorizing?
One way that works in many such inequalities is to try some cases in each section of the reals in question: x=0 and x=-1000 and see which ones work.
 
  • #20
vela said:
The easiest way is using what @BvU has hinted at. Since x2=|x|, after taking the square root of the original inequality, you have |x|>30.
Thank you for clarifying but what is ##\sqrt 900##? Doesn't the inquality ##x^2 > 900## becomes ##|x| > \pm 30## after square rooting?
 
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  • #21
mark2142 said:
Thank you for clarifying but what is ##\sqrt 900##? Doesn't the inquality ##x^2 > 900## becomes ##|x| > \pm 30## after square rooting?
##\sqrt{900} = 30##
If ##x^2 > 900## then ##|x| > 30##

The expression ##\sqrt{900}## represents the principal (i.e., positive) square root of 900. It's a common mistake that we see a lot where people think that the square root of a number is plus/minus some quantity. That is erroneous.
 
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  • #22
Mark44 said:
##\sqrt{900} = 30##
If ##x^2 > 900## then ##|x| > 30##

The expression ##\sqrt{900}## represents the principal (i.e., positive) square root of 900. It's a common mistake that we see a lot where people think that the square root of a number is plus/minus some quantity. That is erroneous.1
https://math.stackexchange.com/a/547831/1109500
Is this what you mean? That we choose ##\sqrt x## to be a positive square root because it helps in solving inequalities and equations. And for that to be consistent all the way we want ##\sqrt {x^2}= |x| ## to be true.
Then it will be like @vela said. ##|x| >30## and by definition of absolute values we have x<-30 or x>+30.
 
  • #23
mark2142 said:
Yes
mark2142 said:
That we choose ##\sqrt x## to be a positive square root because it helps in solving inequalities and equations.
No, we choose ##\sqrt x ## to be the positive square root of x so that the square root operation will be a function. A function has only a single value. However, in more advanced mathematics the concept of multi-valued or vector-valued functions comes up.
mark2142 said:
And for that to be consistent all the way we want ##\sqrt {x^2}= |x| ## to be true.
I don't think that's the right explanation. By definition, |x| equals x if x is is nonnegative, and equals -x if x happens to be negative. With our convention, ##\sqrt{x^2}## has the same value as |x|.
mark2142 said:
Then it will be like @vela said. ##|x| >30## and by definition of absolute values we have x<-30 or x>+30.
 
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  • #24
mark2142 said:
https://math.stackexchange.com/a/547831/1109500
Is this what you mean? That we choose ##\sqrt x## to be a positive square root because it helps in solving inequalities and equations. And for that to be consistent all the way we want ##\sqrt {x^2}= |x| ## to be true.
Then it will be like @vela said. ##|x| >30## and by definition of absolute values we have x<-30 or x>+30.
$$(-31)(-31) = (31)(31) > 900$$$$(-29)(-29) = (29)(29) < 900$$$$-31<-30<-29< 0 < 29 < 30 < 31$$
 
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  • #25
Mark44 said:
don't think that's the right explanation. By definition, |x| equals x if x is is nonnegative, and equals -x if x happens to be negative. With our convention, x2 has the same value as |x|.
When you say ## \sqrt {x^2}= |x|## is this because ## \sqrt y## is taken as positive root where ##y=x^2## ? (And |x| is already positive. So both becomes equal).
Like ## \sqrt {(-2)^2}= \sqrt 4 = +2##
And ## \sqrt {2^2}= \sqrt 4 = +2##
So, ## \sqrt x^2 = |x|## for all x.
 
  • #26
mark2142 said:
So, ## \sqrt x^2 = |x|## for all x.
Exactly.
 
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  • #27
mark2142 said:
When you say ## \sqrt {x^2}= |x|## is this because ## \sqrt y## is taken as positive root where ##y=x^2##?
It doesn't really have anything to do with ##y = x^2##.
If x < 0, then ##\sqrt{x^2} = \sqrt{(-x)^2} = -x##, which is a positive number.
If ##x \ge 0## then ##\sqrt{x^2} = x##, which is nonnegative.
Therefore, ##\sqrt{x^2}## produces the exact same result as |x|.
mark2142 said:
(And |x| is already positive. So both becomes equal).
Like ## \sqrt {(-2)^2}= \sqrt 4 = +2##
And ## \sqrt {2^2}= \sqrt 4 = +2##
So, ## \sqrt x^2 = |x|## for all x.
Yes, with this correction. You wrote ## \sqrt x^2## in the last line. I think I understand what you meant, but that wasn't consistent with the LaTeX that you wrote.
## \sqrt x^2## is rendered as if you had written ##(\sqrt x)^2##, which is defined only for ##x \ge 0## if the intention is for the real-valued square root.
The corrected version of that last line is ##\sqrt{x^2} = |x|##. In other words, you need braces around the x^2 part. So there is a difference between taking the square root first and then squaring the result, versus squaring x first, then taking the square root of that result.
 
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  • #28
PeroK said:
Exactly.
Ok. Great. And the fact that ## (-2)^{2*1/2}= -2## is true but we ignore it and say ## (-2)^{2*1/2}= |-2|=2##. Yes?
 
  • #29
mark2142 said:
Ok. Great. And the fact that ## (-2)^{2*1/2}= -2## is true but we ignore it and say ## (-2)^{2*1/2}= |-2|=2##. Yes?
No.
 
  • #30
mark2142 said:
Ok. Great. And the fact that ## (-2)^{2*1/2}= -2## is true but we ignore it and say ## (-2)^{2*1/2}= |-2|=2##. Yes?
No. The usual rules for exponents don't apply when you raise negative numbers to some power. The rule you apparently are using is ##(a^b)^c = a^{bc}## requires that a > 0.
For the expression you started with, the expressions ##((-2)^2)^{1/2}## and ##(-2)^{2 * 1/2}## produce different results, namely 2 and -2, respectively.
 
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  • #31
mark2142 said:
Ok. Great. And the fact that ## (-2)^{2*1/2}= -2## is true but we ignore it and say ## (-2)^{2*1/2}= |-2|=2##. Yes?
Keep things simple:
$$(x^2)^{1/2} = \sqrt{x^2} = |x|$$$$x^{(2*\frac 1 2)} = x$$
 
  • #32
Mark44 said:
It doesn't really have anything to do with y=x2.
I meant to substitute ##x^2## with ##y## so to make it more clear. Square root of y is defined to be positive root. My explanation seems right.
I am not saying I don’t agree with yours. It’s just I get mine and it’s easy to remember.
 
  • #33
Thank you.
 

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