Is it Possible for x^2 to Exceed 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.

 

[BvU]

How about $|x|$ ?

 

[PeroK]

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 $x \ ? -a$?

 

[Steve4Physics]

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.

 

[FactChecker]

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: $x \lt -\sqrt a$

 

[BvU]

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]

$x > \pm \sqrt a$ 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]

How about $|x|$ ?

What about it?

 

[BvU]

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 ?

$\$

 

[BvU]

What about it?

$x^2 > 900\ \Leftrightarrow \ |x|>\sqrt{900}$

 

[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 $x>30$.

 

[BvU]

I thought maybe there was a fast algebraic method

There is:

$x^2 > 900\ \Leftrightarrow \ |x|>\sqrt{900}$

$\Leftrightarrow \ x>30\ \lor \ -x<-30$

[edit] Oops ! see #16

 

[FactChecker]

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!

 

[DrClaude]

$\Leftrightarrow \ x>30\ \lor \ -x<-30$

$\Leftrightarrow \ x>30\ \lor \ x<-30$

 

[Mark44]

Emphasis added...

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.

 

[vela]

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 $x < -30$.

 

[FactChecker]

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?

 

[Mark44]

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]

$\sqrt{900} = 30$
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.

 

[Mark44]

https://math.stackexchange.com/a/547831/1109500
Is this what you mean?

Yes

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.

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|.

Then it will be like @vela said. $|x| >30$ and by definition of absolute values we have x<-30 or x>+30.

 

[PeroK]

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.

$$(-31)(-31) = (31)(31) > 900$$$$(-29)(-29) = (29)(29) < 900$$$$-31<-30<-29< 0 < 29 < 30 < 31$$

 

[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.

 

[PeroK]

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

Exactly.

 

[Mark44]

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|.

(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.

 

[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?

 

[PeroK]

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.

 

[Mark44]

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.

 

[PeroK]

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.