Problem with simple inequality

[dyn]

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

 

[fresh_42]

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.

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:
$$
9>x^2 \Longrightarrow \begin{cases}
x=\sqrt{x^2}<3=\sqrt{9} &\text{ if }x\geq 0\\ x=-\sqrt{x^2}> -3=-\sqrt{9} &\text{ if }x< 0
\end{cases}
$$

 

[dyn]

Thanks. So . for a first step if i take the square root of both sides of 9 > x² what do i get ?

 

[fresh_42]

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.

 

[jbriggs444]

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.

 

[PeroK]

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|##.

 

[dyn]

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 ?

 

[jbriggs444]

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##.

 

[PeroK]

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.

 

[Mark44]

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##.

 

[statdad]

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

 

[dyn]

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 ?

 

[fresh_42]

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.##

 

[PeroK]

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:
$$x^2 = y^2 \ \Rightarrow \ x = \pm y$$