B Problem with simple inequality

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The discussion centers on the inequality 9 > x² and the confusion surrounding taking the square root of both sides. It clarifies that taking the square root requires careful handling, as it leads to the absolute value, resulting in the correct interpretation of the inequality as -3 < x < 3. The participants emphasize that the square root function only yields the principal root, which is positive, and that the notation ±x is misleading in this context. Ultimately, the correct approach to solving the inequality involves recognizing that taking the square root does not account for both positive and negative solutions unless explicitly stated.
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Hi
If i have the inequality , 9 > x2 then i know the answer is , -3 < x < +3 but my confusion lies in the following ; if i take the square root of both sides of the inequality i get , ± 3 > ±x
Is that correct ? If so , it leads to the following solutions
x < 3 , x > -3 , x < -3 , x >3 ; which i know is wrong but why ?
Thanks
 
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dyn said:
Hi
If i have the inequality , 9 > x2 then i know the answer is , -3 < x < +3 but my confusion lies in the following ; if i take the square root of both sides of the inequality i get , ± 3 > ±x
Is that correct ?
No, it isn't. You cannot write ##\pm a > \pm b.## That makes no sense.
dyn said:
If so , it leads to the following solutions
x < 3 , x > -3 , x < -3 , x >3 ; which i know is wrong but why ?
Thanks

Taking the square root gives you:
$$
9>x^2 \Longrightarrow \begin{cases}
x=\sqrt{x^2}<3=\sqrt{9} &\text{ if }x\geq 0\\ x=-\sqrt{x^2}> -3=-\sqrt{9} &\text{ if }x< 0
\end{cases}
$$
 
Thanks. So . for a first step if i take the square root of both sides of 9 > x2 what do i get ?
 
dyn said:
Thanks. So . for a first step if i take the square root of both sides of 9 > x2 what do i get ?
I would proceed very carefully. Let's see what we have.
\begin{align*}
9 > x^2 &\Longrightarrow 9-x^2 > 0\\
&\Longrightarrow (3-x)\cdot (3+x) > 0\\
&\Longrightarrow \left[3-x > 0 \;\;\;\text{ AND } \;\;\;3+x > 0 \right] \;\;\;\text{ OR } \;\;\;\left[3-x < 0 \;\;\;\text{ AND } \;\;\;3+x < 0 \right]\\
&\Longrightarrow \left[3>x>-3\right] \;\;\;\text{ OR } \;\;\;\left[3<x<-3\right]\\
&\Longrightarrow 3>x>-3 \;\;\;\text{ OR }\;\;\; x\in \emptyset\\
&\Longrightarrow 3>x>-3
\end{align*}
This is the procedure step by step.
 
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dyn said:
Thanks. So . for a first step if i take the square root of both sides of 9 > x2 what do i get ?
If you take the square root of both sides, you will preserve the inequality. This follows since the square root function is monotone increasing. [A monotone increasing function is one in which ##x > y \Rightarrow f(x) > f(y)##]

We must proceed with caution. It is not always the case that ##\sqrt{x^2} = x##. The square root function always returns a positive result. However, there are still two possibilities depending on the sign of x.

If ##x## is positive or zero, then ##\sqrt{x^2} = x## and our inequality becomes ##3 > x##.
If ##x## is negative, then ##\sqrt{x^2} = -x## and our inequality becomes ##3 > -x##.

The latter inequality is easily converted to ##-3 < x## because multiplication by negative one is a monotone decreasing function. It inverts the sense of the inequality when we apply the function to both sides.

The result is the case statement given by @fresh_42 in #2.
 
dyn said:
Thanks. So . for a first step if i take the square root of both sides of 9 > x2 what do i get ?
Note that ##\sqrt{x^2} = |x|##. So, you get ##3 > |x|##.
 
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Thanks everyone. If the inequality is turned into an equals sign and i have 9 = x2 and i take the square root of both sides do i have
±3 = ±x ?
 
dyn said:
Thanks everyone. If the inequality is turned into an equals sign and i have 9 = x2 and i take the square root of both sides do i have
±3 = ±x ?
The square root operation returns the principal square root only. ##\sqrt{9} = 3##. Not ##\pm 3##.
The square root operation returns the principal square root only. ##\sqrt{x^2} = |x|##. Not ##\pm x##.
 
dyn said:
Thanks everyone. If the inequality is turned into an equals sign and i have 9 = x2 and i take the square root of both sides do i have
±3 = ±x ?
No.
 
  • #10
dyn said:
If the inequality is turned into an equals sign and i have 9 = x2 and i take the square root of both sides do i have
±3 = ±x ?
Your answer is "sort of" correct, but your explanation is not.
If ##x^2 = 9## then there are two solutions: x = 3 or x = -3. Your solution, ±3 = ±x, boils down to what I wrote.

Where your explanation is faulty is in thinking that ##\sqrt {x^2} = \pm x##. So taking square roots of the two sides results in ##|x| = 3## or ##x = \pm 3##.
 
  • #11
dyn said:
Thanks everyone. If the inequality is turned into an equals sign and i have 9 = x2 and i take the square root of both sides do i have
±3 = ±x ?
No: if you have

## x^2 = 9##

and you solve it you get

## x = \pm 3 ##
If you were to, in isolation, compute $$\sqrt{9}$$ you would get $$3$$, not $$\pm 3$$. That is becuase

- when you solve the equation you are looking for all possible solutions
- the ``surd operator'' $$\sqrt{\hphantom{9}}$$ returns, by definition, only the positive square root
 
  • #12
So if i have x2 = 9 and i take the square root of both sides i get x = 3 which is obviously not the full solution, so taking the square root of both sides does not give the full solution ?
 
  • #13
dyn said:
So if i have x2 = 9 and i take the square root of both sides i get x = 3 which is obviously not the full solution, so taking the square root of both sides does not give the full solution ?
##x^2=9## becomes
$$
0=x^2-9=(x+3)\cdot (x-3) \quad \Longrightarrow \quad x=-3 \;\;\;\text{ OR } \;\;\;x=3
$$
Full solution. No need to bother the square root.

The square root is only a function for ##x\geq 0## and it is defined as ##\sqrt{.}\, : \,\mathbb{R}^+_0 \longrightarrow \mathbb{R}^+_0.## There is another, related, however, different function ##-\sqrt{.}\, : \,\mathbb{R}^+_0 \longrightarrow \mathbb{R}^-_0.## You can talk about either function, but not about them as it was only one function. It is not. ##(x,\pm\sqrt{x})=\{(x,\sqrt{x})\,|\,x\geq 0\}\cup \{(x,-\sqrt{x})\,|\,x\geq 0\}## is a relation, not a function.

Every function is a relation, but not every relation is a function. And since your notation ##\pm \sqrt{x^2}## refers to the relation with that ##\pm## you cannot pretend to apply a function on both sides of ##9=x^2.## What you can do is apply the function ##\sqrt{.}## and get ##\sqrt{9}=3=\sqrt{x^2}=x## like you would apply the function, e.g. times ##4## and get ##36=4x^2.## You can also apply the function ##-\sqrt{.}## on both sides and get ##-\sqrt{9}=-3=x.## But how do we know that we cannot find even more solutions if we only applied other functions, too? That's why ##0=9-x^2=(x-3)(x+3)## is the correct handling. This allows only the two solutions ##x=\pm 3.##
 
Last edited:
  • #14
dyn said:
So if i have x2 = 9 and i take the square root of both sides i get x = 3 which is obviously not the full solution, so taking the square root of both sides does not give the full solution ?
No, that's why you often have ##\pm \sqrt{n}##. E.g. in the quadratic formula. In general:
$$x^2 = y^2 \ \Rightarrow \ x = \pm y$$
 

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