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So I've had some trouble understanding what happens when you take the square roots of an inequality.

Am I correct in saying that if

x^2 > n

then

-squareroot(n) > x > +squareroot(n)

and if

x^2 < n

then

-squareroot(n) > x < +squareroot(n)

??

Jean-Claude Arbaut
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Jin
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3 Answers3

1

When you take the square root you have the absolute value, so $$x^2 < n \Rightarrow \sqrt{x^2} < \sqrt{n} \\ \Rightarrow |x|<\sqrt{n}$$ That then implies that $-\sqrt{n}<x<\sqrt{n}$

For the other case we have $$x^2 > n \Rightarrow \sqrt{x^2} > \sqrt{n} \\ \Rightarrow |x|> \sqrt{n}$$ which then implies $x>\sqrt{n}$ OR $x<-\sqrt{n}$

Sheheryar Zaidi
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1

You can also draw a picture:

enter image description here

The parabola represents $x^2$ and the horizontal line represents $n$ (assumed to be positive in this picture).

Where do the blue parabola and the purple horizontal line intersect? From that, what can you say about the values of $x$ such that $x^2>n$? What about $x^2<n$?

Kim Jong Un
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0

No, it's incorrect, for example for $n=4$ and $x=5$ (for first) and $x=1$ (for second). It should be:

If $x^2>n$, then $x>\sqrt{n}$ or $x<-\sqrt{n}$.

If $x^2<n$, then $-\sqrt{n}<x$ and $x<\sqrt{n}$

agha
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  • Yes, it's incorrect, because $-2<5$. – agha Sep 01 '14 at 20:22
  • How is the first one incorrect for 4? I don't get it. Let's do an example with n = 4, so that x^2 > 4 then

    -squareroot(4) > x > +squareroot(4) .... then x must be bigger than 2 (so for example 3^2 is bigger than 4), and it must be smaller than -2, so (-3)^2 is also bigger than 4. Why is this wrong?

     – Jin Sep 01 '14 at 20:22
  • Note that negative number is always smaller than positive. $-\sqrt{n}$ is always negative. – agha Sep 01 '14 at 20:25
  • Look at the graph posted above. It clearly indicates that for x'es smaller than -2 (i.e. -3, -4, so on) and for x'es bigger than 2 (3,4, so on), our inequality is satisfied..... so x > squareroot(4)(3,4, so on) or x < -squareroot(2) (-3,-4, so on).... what's wrong about this? Please explain, because I must be missing something very fundamental here. – Jin Sep 01 '14 at 20:31
  • But for $x<-2$ you have $x<\sqrt{n}$.... Oh course, when $x>\sqrt{n}$ you have $x^2>n$, but if $x>-\sqrt{n}$ not necessery. But if $x<\sqrt{n}$ you have $x^2>n$. – agha Sep 01 '14 at 20:35