Quadratic equation: Which way is correct? pic1 or pic2?

[Callmelucky]

Homework Statement

I think that the way in pic 1 is right because of the property written next to the procedure but the professor who posts videos on youtube solved it the way as written in pic 2

Relevant Equations

ax^2 + bx + c = 0, |x|>=0

I am a bit confused, so if anyone can explain to me which way is right I would be very thankful.

I think that the way in pic 1 is right because of the properties written next to the procedure but the professor who posts videos on youtube solved it the way as written in pic 2 where he didn't consider possible cases of the absolute value of the expression. He just removed absolute values and set the expression equal to 1/2 and -1/2. Solutions in both cases are the same since the solutions in 1st pic match with conditions but, what if they didn't match? That is what bothers me.

Thank you.

 

[DaveE]

OK, a few observations:

1) You never actually presented the problem you are asking about. You'll get more and better answers if you make your posts easy to read. I usually skip posts like this because of the extra work and high probability that I won't really understand what question is being asked.

2) Both pages arrived at the same answer. So why does one method have to be "right". There are often different ways of solving the same problem.

3) If their answers weren't the same then it means one of them is incorrect or incomplete. That usually has nothing to do with the method. It's usually just an algebraic error. So you would read through each step and see if you agree with their reasoning. You'll find one has an error and then fix it to get the correct answer.

 

[topsquark]

Homework Statement:: I think that the way in pic 1 is right because of the property written next to the procedure but the professor who posts videos on youtube solved it the way as written in pic 2
Relevant Equations:: ax^2 + bx + c = 0, |x|>=0

I am a bit confused, so if anyone can explain to me which way is right I would be very thankful.

I think that the way in pic 1 is right because of the properties written next to the procedure but the professor who posts videos on youtube solved it the way as written in pic 2 where he didn't consider possible cases of the absolute value of the expression. He just removed absolute values and set the expression equal to 1/2 and -1/2. Solutions in both cases are the same since the solutions in 1st pic match with conditions but, what if they weren't the same? That is what bothers me.

Thank you.

First, why do you have $x \geq 0$ in the problem statement? As it happens, neither solution satisfies this!

The first solution uses < > symbols. I have no idea why. This is an equation, not an inequality. Once they were changed to = the solution is fine.

The second solution does the same thing, except instead of saying
$| 3x + 2 | = \dfrac{1}{2}$

it uses
$3x + 2 = -1/2$ or $3x + 2 = 1/2$.

Both methods say the same thing.

-Dan

 

[mjc123]

Prof's way is right. It covers all the possibilities.

I'm not sure what you're doing in pic 1, but you are being inconsistent about square roots. You can't say
√(x²) = |x| and √(1/4) = ±1/2
And what you're doing on the next few lines makes no sense to me.
The other way is simpler and better.

 

[Frabjous]

They are both correct. You have
f(x)²=C²
which gives
±f(x)=±C
which gives 4 possibilities
f(x)=C
f(x)=-C
-f(x)=C
-f(x)=-C
Notice that 1st and 4th differ by a sign and give indentical solutions
Similarly for the 2nd and 3rd.
You used 1st and 3rd. The prof used 1st and 2nd.

 

[Callmelucky]

Prof's way is right. It covers all the possibilities.

I'm not sure what you're doing in pic 1, but you are being inconsistent about square roots. You can't say
√(x²) = |x| and √(1/4) = ±1/2
And what you're doing on the next few lines makes no sense to me.
The other way is simpler and better.

I don't know. In the textbook, it's written that √(x^2) = |x|, and then some pages after √(1/4) = ±1/2.
I guess that is what bothers me. I don't understand.
While I was doing absolute value equations the first rule was introduced and now when I am doing quadratic equations I was told that it can have two possible solutions. Both cases make sense. So I don't know.

 

[Callmelucky]

They are both correct. You have
f(x)²=C²
which gives
±f(x)=±C
which gives 4 possibilities
f(x)=C
f(x)=-C
-f(x)=C
-f(x)=-C
Notice that 1st and 4th differ by a sign and give indentical solutions
Similarly for the 2nd and 3rd.
You used 1st and 3rd. The prof used 1st and 2nd.

I know but, what if conditions weren't matched? I guess that can happen.

 

[Callmelucky]

OK, a few observations:

1) You never actually presented the problem you are asking about. You'll get more and better answers if you make your posts easy to read. I usually skip posts like this because of the extra work and high probability that I won't really understand what question is being asked.

2) Both pages arrived at the same answer. So why does one method have to be "right". There are often different ways of solving the same problem.

3) If their answers weren't the same then it means one of them is incorrect or incomplete. That usually has nothing to do with the method. It's usually just an algebraic error. So you would read through each step and see if you agree with their reasoning. You'll find one has an error and then fix it to get the correct answer.

I don't know why are pics blurry, on my phone they are crystal clear. I have written problem from notebook to paper to make it more clear but I guess it didn't help much. I will learn latex asap.

 

[Mark44]

Homework Statement:: I think that the way in pic 1 is right because of the property written next to the procedure but the professor who posts videos on youtube solved it the way as written in pic 2

We have a policy that discourages posting images of posters' work, as they are usually difficult to read. See this post -- https://www.physicsforums.com/threads/homework-help-guidelines-for-students-and-helpers.686783/.
You will be lucky if your next homework thread that includes an image doesn't get locked. Take a few minutes and learn how to use LaTeX from our tutorial whose link is at the lower left corner of the input pane.

Relevant Equations:: ax^2 + bx + c = 0, |x|>=0

The restriction $|x| \ge 0$ is no restriction at all. It will be true for any real number x.

 

[Mark44]

It is tricky when you are learning this stuff. When you take a square root there is always a positive and negative value unless it is zero. Sometime, by convention, you assume that positive value is wanted.

No.
Although it is true that every positive real number has two square roots, the square root of a positive real number is by convention positive. The symbol $\sqrt x$ represents the positive number y such that $y^2 = x$.

It is incorrect to say, for example, that $\sqrt 4 = \pm 2$.

 

[Callmelucky]

No.
Although it is true that every positive real number has two square roots, the square root of a positive real number is by convention positive. The symbol $\sqrt x$ represents the positive number y such that $y^2 = x$.

It is incorrect to say, for example, that $\sqrt 4 = \pm 2$.

Why then do we have +1/2 and -1/2 as a result of sqrt(1/4)? I mean if sqrt of positive is positive?

 

[Mark44]

Why then do we have +1/2 and -1/2 as a result of sqrt(1/4)?

We don't.
$\sqrt{\frac 1 4} = \frac 1 2$.

However, the equation $x^2 = \frac 1 4$ has two solutions: x = 1/2 or x = -1/2. They are not obtained by taking the square root of both sides of this equation.

 

[Callmelucky]

We don't.
$\sqrt{\frac 1 4} = \frac 1 2$.

Well, that is exactly what the professor did. Not that but he wrote that sqrt(1/4) = +1/2 and -1/2

 

[Frabjous]

No.
Although it is true that every positive real number has two square roots, the square root of a positive real number is by convention positive. The symbol $\sqrt x$ represents the positive number y such that $y^2 = x$.

It is incorrect to say, for example, that $\sqrt 4 = \pm 2$.

By your interpretation, method 1 is correct and method 2 is wrong. I bet that if you were solving the problem, you would do method 2.

 

[Frabjous]

It is all a convention thing. As I pointed out in post 5, there are 4 possible combinations of which only 2 are unique. Your method follows the positive root rule, but in practice most people use the method your prof uses.

 

[Callmelucky]

We don't.
$\sqrt{\frac 1 4} = \frac 1 2$.

However, the equation $x^2 = \frac 1 4$ has two solutions: x = 1/2 or x = -1/2. They are not obtained by taking the square root of both sides of this equation.

and in my textbook it's stated that x^n=a, when n is even number and a>0 x= +,- sqrt(a)
Just saw this "However, the equation $x^2 = \frac 1 4$ has two solutions: x = 1/2 or x = -1/2. They are not obtained by taking the square root of both sides of this equation."
And it makes sense now.

 

[Mark44]

By your interpretation, method 1 is correct and method 2 is wrong.

If, by method 1, you mean the work in 1.jpeg, that image has several problems.

$|3x + 2| = \pm \frac 1 2$

and

$\sqrt{\frac 1 4} = \pm \frac 1 2$

Both are incorrect. The first because the absolute value always is nonnegative. The second because, as I already said, the square root of a positive number is always positive.

It is all a convention thing. As I pointed out in post 5, there are 4 possible combinations of which only 2 are unique. Your method follows the positive root rule, but in practice most people use the method your prof uses.

Half of the equations you wrote are redundant. If $f(x)^2 = C^2$, then $f(x) = \pm C$. That's all you need to say.

The "positive root rule" is the one that all people knowledgeable with mathematics use. No mathematician would write $\sqrt 4 = \pm 2$, for example.

and in my textbook it's stated that x^n=a, when n is even number x= +,- sqrt(a)

No, it doesn't say that, I guarantee you. It might say that if $x^2 = a$, then $x = \pm \sqrt a$, provided that a > 0, but it's not true that if $x^4 = a$, then $x = \pm \sqrt a$.

 

[Mark44]

Well, that is exactly what the professor did. Not that but he wrote that sqrt(1/4) = +1/2 and -1/2

IF said professor wrote $\sqrt{\frac 1 4} = \pm \frac 1 2$, then he/she is flat wrong.
OTOH, if what was written was $x = \pm \frac 1 2$, then that is correct. I explained why in post #13.

 

[Callmelucky]

I finally get it :D

I appreciate everyone's help, but this comment "However, the equation x^2=1/4 has two solutions: x = 1/2 or x = -1/2. They are not obtained by taking the square root of both sides of this equation." just made my week.

Thank you everyone.

 

[Frabjous]

Half of the equations you wrote are redundant. If f(x)2=C2, then f(x)=±C. That's all you need to say.

Yes, which is why I said that only 2 are unique. I wrote them out so that the OP could see all of them. Following the convention, shouldn’t the answer be ±f(x)=C which sort of makes my point.

edit: C>0

 

[Mark44]

Following the convention, shouldn’t the answer be ±f(x)=C which sort of makes my point.

There's no difference between what you wrote here and $f(x) = \pm C$. Most often you would see it as I wrote it, with $\pm$ on the right side.

 

[SammyS]

OTOH, if what was written was $x^2 = \pm \frac 1 2$, then that is correct. I explained why in post #13.

Hey Mark.
I think you have a typo there. $x$ shouldn't be squared.

 

[Frabjous]

There's no difference between what you wrote here and $f(x) = \pm C$. Most often you would see it as I wrote it, with $\pm$ on the right side.

Which sounds like a different convention which I believe was the source of the OP’s confusion.

 

[Callmelucky]

IF said professor wrote $\sqrt{\frac 1 4} = \pm \frac 1 2$, then he/she is flat wrong.
OTOH, if what was written was $x = \pm \frac 1 2$, then that is correct. I explained why in post #13.

Actually, he did say, that x= +, - sqrt(1/4), I was wrong.

 

[Callmelucky]

Which sounds like a different convention which I believe was the source of the OP’s confusion.

There were a few things that were unclear to me, but I think I got them now. Thanks everyone.

 

[Mark44]

This thread reminded me of an instructor at the college I taught at for 18 years. This was a guy who was tenured and had been at the college for many years. He would teach his students stuff like this:
If $x^2 = 9$, then just apply the square root operator to "cancel" (as he put it) the exponent and get $x = 3$.

All of the other math instructors in the department (about a dozen) were appalled at the ignorance his over-simplification showed, as it completely ignored that fact that x = -3 was also a solution. His erroneous work wasn't limited to just problems like this. He would "cancel" all sorts of other operations without apparently realizing that what he was doing wasn't valid at all. Furthermore, no amount of reasoning from the rest of us could sway him from his wrongheadedness.

The quote from Albert Einstein seems especially apt: "Make things as simple as possible, but no simpler."

 

[Mark44]

I think you have a typo there. x shouldn't be squared.

Yep, typo. I've fixed it now.

 

[Mark44]

Following the convention, shouldn’t the answer be ±f(x)=C which sort of makes my point.

edit: C>0

$\pm f(x) = C$ is equivalent to $f(x) = \pm C$. Both equations say exactly the same thing. Doesn't matter whether C is positive or negative.

Which sounds like a different convention which I believe was the source of the OP’s confusion.

I don't think that's where the OP was confused. In any case, the convention that is recognized around the world, is that the square root of a positive real number is positive, not negative, and not two values.

 

[Frabjous]

This thread reminded me of an instructor at the college I taught at for 18 years. This was a guy who was tenured and had been at the college for many years. He would teach his students stuff like this:
If $x^2 = 9$, then just apply the square root operator to "cancel" (as he put it) the exponent and get $x = 3$.

All of the other math instructors in the department (about a dozen) were appalled at the ignorance his over-simplification showed, as it completely ignored that fact that x = -3 was also a solution. His erroneous work wasn't limited to just problems like this. He would "cancel" all sorts of other operations without apparently realizing that what he was doing wasn't valid at all. Furthermore, no amount of reasoning from the rest of us could sway him from his wrongheadedness.

The quote from Albert Einstein seems especially apt: "Make things as simple as possible, but no simpler."

How insulting. I’ll stop watching this thread now.

 

[Mark44]

How insulting.

Why?
It wasn't aimed at you or the OP. It was about someone (the instructor at my college) who wasn't just misinformed but was willfully teaching his students stuff that wasn't true, and that he should have realized was untrue.

 

[anuttarasammyak]

As the convention
$$\sqrt{}$$
denotes nonnegative square root , e.g. 2 out of 2,-2 which are square roots of 4.

That is OK but I am puzzled in using this symbol for negative or complex numbers.
For an example
$$\sqrt{-1}=i$$
Why not -i ? What is the convention ? I suppose it is "nonnegative on pure imaginary axis". Is it right?
Square roots of i are $e^{\pi/4\ i},e^{5\pi/4\ i}$. Which is $\sqrt{i}$ ?

 

[Mark44]

That is OK but I am puzzled in using this symbol for negative or complex numbers.
For an example
$$\sqrt{-1}=i$$
Why not -i ? What is the convention ? I suppose it is "nonnegative on pure imaginary axis". Is it right?
Square roots of i are $e^{\pi/4\ i},e^{5\pi/4\ i}$. Which is $\sqrt{i}$ ?

"Nonnegative on pure imaginary axis" seems to be the convention. For your second question, several web sites I looked at (search on "principal square root of a complex number") define the principal square root of a complex number as the root with a positive imaginary part. It's important to note that the imaginary part is the coefficient (a real number) of i. This would make $e^{\pi/4 i}$ the principal square root of i.

 

[anuttarasammyak]

Thank you so much @Mark44 .

So we may summarize the convention of $\sqrt{}$ that we choose from square roots
#1 nonnegative one on imaginary axis
If imaginary part are zero,
#2 nonnegative one on real axis.
Or in a word it has phase angle of $\phi[0,\pi)$.

$$\sqrt{e^{i5\pi/4}}=e^{i5\pi/8}$$
whose real part is negative and imaginary part is positive. Imaginary axis prevails.