# Inequalities: x^2 > 900

mark2142
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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• vela
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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##?

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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.
##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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• mark2142
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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##

• mark2142
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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 ...

##\ ##

• mark2142
mark2142
##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##.)

mark2142
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! )

mark2142

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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 ? ##\ ##

• MatinSAR, scottdave, PeroK and 1 other person
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##x^2 > 900\ \Leftrightarrow \ |x|>\sqrt{900}##

• MatinSAR
mark2142
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##.

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I thought maybe there was a fast algebraic method
There is:
##x^2 > 900\ \Leftrightarrow \ |x|>\sqrt{900}##
##\Leftrightarrow \ x>30\ \lor \ -x<-30 ##

 Oops ! see #16

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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.

mark2142
Not really. The best (only?) way is to consider two cases and see where each case takes you.
Ok! Thanks everyone. Bye!

Mentor
##\Leftrightarrow \ x>30\ \lor \ -x<-30 ##
##\Leftrightarrow \ x>30\ \lor \ x<-30 ##

• Mark44
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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.

• • scottdave and mark2142
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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##.

• mark2142 and scottdave
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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.

mark2142
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?

• PeroK
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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.

• mark2142 and vela
mark2142
##\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.

Mentor
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.

• MatinSAR, vela and SammyS
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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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mark2142
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.

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So, ## \sqrt x^2 = |x|## for all x.
Exactly.

• mark2142
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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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mark2142
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?

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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.

Mentor
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.

• • mark2142 and PeroK
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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:
$$(x^2)^{1/2} = \sqrt{x^2} = |x|$$$$x^{(2*\frac 1 2)} = x$$

mark2142
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.

mark2142
Thank you.