radical problem

[cschulz57919]

directions:solve.

Problem: [(Z/4)-5]^2/3=1/25

Obviously I'm supposed to solve for z. so I squared inside the brackets

(cube root of)(z²/16 - 25)=1/25

you cube both sides to get rid of the radical

z²/16-25=1/15625

add 25 to both sides

z²/16=390626/15625

times 16 to both sides

z²=400.001024

square root both sides

20.0000256=z

plug it in to original

[(20.0000256²/4²)-5²]^1/3=2.714420512

... thats not 1/25

help... where did I go wrong?

 

[tkhunny]

Right up front.

(2 + 3)^2 = 5^2 = 25

Your method

(2 + 3)^2 = (2^2 + 3^2) = 4 + 9 = 13!! Whoops! Something wrong with that.

Correctly

(2 + 3)^2 = (2 + 3)*(2 + 3) = 4 + 6 + 6 + 9 = 10 + 15 = 25 -- That's better.

 

[cschulz57919]

reply to JeffM

THANK YOU!!!!!!!!

That helped so much. especially cuz i could see it in normal writing.

 

[lookagain]

JeffM,

your method got bogged down with expanding to a quadratic and using the
Quadratic Formula and numbers with lots of decimal digits.


Denis,

your method is missing one of the solutions.


-------------------------------------------------


Let's use the root method:


\(\displaystyle (z/4 - 5)^{2/3} = \dfrac{1}{25}\)


Cube each side:


\(\displaystyle (z/4 - 5)^2 = \dfrac{1}{25^3}\)


\(\displaystyle (z/4 - 5)^2 = \dfrac{1}{5^6}\)


Take the square roots of each side:


\(\displaystyle z/4 - 5 = \pm\dfrac{1}{5^3}\)


\(\displaystyle z/4 - 5 = \pm\dfrac{1}{125}\)


\(\displaystyle z/4 = 5 \pm 0.008\)


\(\displaystyle z = 4(5 \pm 0.008)\)


\(\displaystyle z = 20 \pm 0.032\)



\(\displaystyle z = 19.968 \ \ \ or \ \ z = 20.032\)

 

[soroban]

Hello, cschulz57919!

\(\displaystyle \text{Solve for }z\!:\;\;\left(\dfrac{z}{4}-5\right)^{\frac{2}{3}} \:=\:\dfrac{1}{25}\)

Raise both sides to the power \(\displaystyle \frac{3}{2}\)

. . . \(\displaystyle \bigg[\left(\dfrac{z}{4} - 5\right)^{\frac{2}{3}}\bigg]^{\frac{3}{2}} \;=\;\left(\dfrac{1}{25}\right)^{\frac{3}{2}} \;=\;\left(\dfrac{1}{5^2}\right)^{\frac{3}{2}} \;=\;\dfrac{1}{5^3}\)

. . . \(\displaystyle \dfrac{z}{4} - 5 \;=\;\dfrac{1}{125} \quad\Rightarrow\quad \dfrac{z}{4} \;=\;5+ \dfrac{1}{125} \;=\;\frac{626}{125}\)

. . . \(\displaystyle z \;=\;\dfrac{2504}{125} \;=\;20.032\)


Edit: corrected my typo.
. . . .Some "typo", eh? .I solved the wrong problem!
. . . .Thanks for the heads-up, Subhotosh!
.

 

[Deleted member 4993]

Soroban ,

There is a small typo in your response

Raise both sides to the power \(\displaystyle \frac{3}{2}\)

. . . \(\displaystyle \bigg[\left(\dfrac{z}{4} - 5\right)^{\frac{2}{3}}\bigg]^{\frac{3}{2}} \;=\;[\dfrac{1}{25}]^{\frac{3}{2}}\)[/QUOTE]

 

[lookagain]

Hello, cschulz57919!


Raise both sides to the power \(\displaystyle \frac{3}{2}\)

. . . \(\displaystyle \bigg[\left(\dfrac{z}{4} - 5\right)^{\frac{2}{3}}\bigg]^{\frac{3}{2}} \;=\;\left(\dfrac{1}{25}\right)^{\frac{3}{2}} \;=\;\left(\dfrac{1}{5^2}\right)^{\frac{3}{2}} \;=\;\dfrac{1}{5^3}\)

. . . \(\displaystyle \dfrac{z}{4} - 5 \;=\;\dfrac{1}{125} \quad\Rightarrow\quad \dfrac{z}{4} \;=\;5+ \dfrac{1}{125} \;=\;\frac{626}{125}\)

. . . \(\displaystyle z \;=\;\dfrac{2504}{125} \;=\;20.032\)


Edit: corrected my typo.
. . . .Some "typo", eh? .I solved the wrong problem!
. . . .Thanks for the heads-up, Subhotosh!
.

soroban,

as was with Denis, you are missing one of the solutions.

Please see the end of post #3 (JeffM) or the end of mine (post #6).


Anyone here can check the other value, z = 19.968, in the
original equation and see that it checks.

 

[lookagain]

... can check the other value, z = 19.968,
in the original equation and see that it checks.

I was looking into the following for a visual demonstration:

My TI-83 Plus graphics calculator

Suggested settings:


\(\displaystyle y_1 = \) (X/4 - 5)^(2/3)


\(\displaystyle y_2 = \)1/25


Window:

Xmin = 19.9
Xmax = 20.1

Ymin = 0
Ymax = .1


Press "GRAPH."


[2nd] [TRACE] [5: intersect]


And continue...

 

[lookagain]

Original equation: (z/4 - 5)^(2/3) = 1/25
z = 19.968
(19.968/4 - 5)^(2/3) = 1/25
(4.992 - 5)^(2/3) = 1/25
(-.008)^(2/3) = 1/25
??????

Either way, by the Order of Operations, we have:



\(\displaystyle (\sqrt[3]{-0.008})^2 \ = \ (-0.2)^2 \ = \ 0.04 \ = \ 1/25\)


\(\displaystyle \sqrt[3]{(-0.008)^2} \ = \sqrt[3]{0.000064} \ = \ 0.04 \ = 1/25\)


_____________________________________________

(-.008)^2^(1/3) = 1/25 ; yeah!

\(\displaystyle * * \)(-.008)^(1/3)^2 = "imaginary" ; oh no

Only speaking as far as the boxed quote portion immediately above,

the quantity in line ** is not "imaginary" (unless you don't think so
but think that certain others would claim it to be)

And, you are missing grouping symbols, as is often told to
students to use (by you and other veteran users), to eliminate
possible ambiguity.

Instead, I would have typed this in non-Latex:


[(-.008)^2]^(1/3) = 1/25


[(-.008)^(1/3)]^2 = 1/25



_________________________________________________

Solve: x^(2/4) = 5
"Let's use the root method":

Take each side to 4th power:
x^2 = 5^4

Take square root of each side:
x^1 = +- 5^2
x = +- 25

Suppose I were to check x = -25 to see if it satisfies the given equation in your example:


\(\displaystyle (-25)^{2/4} \ = \ (\sqrt[4]{-25})^2 \ = \ (\sqrt{\sqrt{-25}})^2 \ = \ \sqrt{-25} \ = \ 5i \ \ne 5 \ = \ (25)^{2/4}\)

 

[daon2]

http://www.wolframalpha.com/input/?i=[(19.968%2F4)-5]^(2%2F3)-1%2F25

There is only one solution.

For futher verification, one can plot the graph of y=(x/4-5)^(2/3)-1/25 and see that it cross the x-axis only once.

 

[daon2]

When \(\displaystyle x \ge 0\) you may happily use rules such as \(\displaystyle x^{a/b} = (x^{a})^{1/b} = (x^{1/b})^a\). But problems happen when \(\displaystyle x<0\). For example, \(\displaystyle (-1)^{2/3} =e^{2\pi i/3}\) is complex and is not equal to \(\displaystyle ((-1)^2)^{1/3}\).

I should also be mentioned that \(\displaystyle z^{ab} \neq (z^a)^b\) when \(\displaystyle z\) is not real (in general).

 

[pka]

I am now honestly confused myself. I thought the discussion was about how best to explicate an answer. Now it appears that I gave the wrong answer. Obviously explicating a wrong answer is not a good idea. Two solutions. I am sincerely interested in understanding why that logic is incorrect?

@JeffM, I know of several different authors of widely used textbooks who disagree with one another on this point.
Consider \(\displaystyle f(x) = \sqrt[3]{{\left( {\frac{x}{4} - 5} \right)^2 }} - 0.04\;\& \,g(x) = \left( {\frac{x}{4} - 5} \right)^{\frac{2}{3}} - 0.04\)
I know of at least two computer algebra systems that both treat \(\displaystyle f(19.968)~\&~g(19.968)\) as two different values: \(\displaystyle f\) gives 0 but \(\displaystyle g\) gives a complex value.

So using a CAS written in the \(\displaystyle f\) form it has two roots, but only one root in the \(\displaystyle g\) form.

The author Earl Swokowski insists that this rule must hold:
\(\displaystyle x^{\frac{2}{3}}=\left(x^2\right)^{\frac{1}{3}}=\left(x^{\frac{1}{3}}\right)^2\).
Following that convention, there are definitely two roots.
But many calculators and CAS use complex exponentiation to evaluate fractional exponents.

 

[lookagain]

http://www.wolframalpha.com/input/?i=[(19.968%2F4)-5]^(2%2F3)-1%2F25

There is only one solution. \(\displaystyle No.\)


For futher verification, one can plot the graph of y=(x/4-5)^(2/3)-1/25
and see that it cross the x-axis only once.

\(\displaystyle \text{It does not cross the x-axis only once.}\)


Let's zoom in on the cusp as it goes toward and away from the x-axis.


For one of the sites, type:

www.quickmath.com


Click on "Plot" of the "Equations" section.


In the far left space of the yellow rectangle, type:

"y = (x/4 - 5)^(2/3) - 0.04"


Suggestions for x-values and for y-values:


\(\displaystyle x = \boxed{19.9} \ \ to \ \ x = \boxed{20.1}\)

\(\displaystyle . \ \ \ \ \ \ \ \ \ \ \ \ and\)

\(\displaystyle y = \boxed{-.1} \ \ to \ \ y = \boxed{.1}\)



Click "Plot" to see the graph.



------------------------------------------------------------------


\(\displaystyle y = (\frac{x}{4} - 5)^{\frac{2}{3}} \ \ is \ a \ transformation \ of \ \ y \ = \ x^{\frac{2}{3}}.\)


The former is equivalent to:


\(\displaystyle y = [\frac{1}{4}(x - 20)]^{\frac{2}{3}}\)


This has both a horizontal shift to the right 20 units and a

compression factor of \(\displaystyle \frac{1}{4}\) of the latter equation.



---------------------------------------------------


Edit: I looked at some information on principal
cube roots here:

http://en.wikipedia.org/wiki/Cube_root

 

[daon2]

Look again, the issue is which root do we want. If one uses pka's identity then all is fine. But the principal root of a negative number is by convention complex (non-real).

 

[daon2]

Maybe this will help. Assuming \(\displaystyle a\) is an integer:

\(\displaystyle x^{\frac{1}{a}} = e^{1/a \ln(x)} = e^{1/a(\ln|x| + i\theta)}\)

For the principal answer, it is the \(\displaystyle \theta\) which corresponds to the complex number \(\displaystyle x\) with \(\displaystyle -\pi < \theta \le \pi\).

If \(\displaystyle x \ge 0\) is real, then \(\displaystyle \theta = 0\) and it simplifies to \(\displaystyle |x|^{1/a}=\sqrt[a]{x}\) - where here I mean the unique positive real root.

If \(\displaystyle x< 0\) is real then \(\displaystyle \theta=\pi\) and it simplifies to \(\displaystyle (|x|^{1/a})e^{i\pi/a}=\sqrt[a]{|x|}e^{i\pi/a}\)

As always there will be \(\displaystyle a\) choices for "the \(\displaystyle a^{th}\) root of \(\displaystyle x\)." So unless some convention is placed, the notation would be ambiguous.

For complex numbers, the absolute values would of course be norms.

 

[pka]

I have never understood what irrational exponents represent operationally, and I am just as clueless about complex exponents. But to make sure I understand what you have told me, I am going to play it back using my own words.
With respect to z^(m / n), where z is a non-negative real and m and n are integers (n not zero), all mathematicians agree that it is convenient to define things so that
\(\displaystyle z^{(m / n)} = \sqrt[n]{z^m} = (\sqrt[n]{z}\ )^m\)
If, however, z is a negative real or a complex number that is not real, some mathematicians believe that it is convenient to retain that convention whereas others do not. Without pretending to understand the grounds for the disagreement, have I grasped the scope and nature of the disagreement itself?

I am sorry if I lead you to think that mathematicians disagree on the meaning of roots. That is not the case. The argument is about notation and how we teach mathematics. Swokowski would have argued that his targeted audience is undergraduates in North America, where undergraduate mathematics is mainly done with real numbers. He also says that in \(\displaystyle z^{(m / n)} = \sqrt[n]{z^m} = (\sqrt[n]{z}\ )^m\) the integers \(\displaystyle m~\&~n\) are relatively prime.
Moreover, there is a growing number of mathematicians who think that the teaching of complex numbers should be more geometrical. Many think that no radical sign should be used with complex numbers.

I have never understood what irrational exponents represent operationally…

In the real numbers we want \(\displaystyle a^x,~a>0\) to be a continuous function. One way to construct the real numbers is by way of the rational numbers. Every irrational number is the limit of a sequence of rational numbers. Because we do have an operational understanding of rational exponents therefore we define irrational exponents as a limit of rational exponents.
That may not help you much. But it is what it is.

 

[lookagain]

Yep; that does show a crossing of the x-axis.

But these do not:
"y = ((x/4 - 5)^2)^(1/3) - 0.04"
or
"y = ((x/4 - 5)^(1/3))^2 - 0.04"

I do not think I understand you.

At this www.quickmath.com site, if I have my settings, for instance, on the equivalent to

x min = 19.5, x max = 20.5, y min = -.1, and y max = .1, then regardless of each of the

following forms, it shows the same graph. And each graph shows (the same) two x-intercepts.



y = (x/4 - 5)^(2/3) - 0.04


y = ((x/4 - 5)^2)^(1/3) - 0.04


y = ((x/4 - 5)^(1/3))^2 - 0.04

 

[pka]

That leaves me one last question on the notational issue. Is there dispute or agreement on the next proposition:
\(\displaystyle If\ z\ has\ a\ not\ real\ n^{th}\ root\ and\ y\ is\ an\ n^{th}\ real\ root\ of\ z,\) \(\displaystyle then\ y \neq \sqrt[n]{z} = z^{(1/n)}.\)

I happen to be squarely in the camp of those who think that we ought not apply any radical notation to complex numbers. Therefore, seeing \(\displaystyle \sqrt[n]{{z^m }}\) I assume that \(\displaystyle z\) is a real number.
Here are some examples that give some idea about the confusion when going back an forth between radical notation and fractional exponents.
\(\displaystyle \sqrt{{x^4 }}=x^2\) but \(\displaystyle \sqrt{{x^2 }}\ne x\)

Say that \(\displaystyle n~\&~m\) have no common factors other than 1.
If \(\displaystyle n\) is an odd integer then \(\displaystyle \sqrt[n]{{x^m }}\) is defined for any \(\displaystyle x\) and the value is non-negative if \(\displaystyle m\) even, negative if \(\displaystyle m\) is odd with \(\displaystyle x<0\).

 

[daon2]

daon

Thank you for trying to further my mathematical education, which sort of came to a screaching halt with the rational numbers. In my one-semester course on abstract algebra, we studied extending natural numbers to integers and rationals, but were stopped dead at reals with the comment that we were too uneducated (perhaps too dense) to grasp them. I have been with Bishop Berkeley all these years: the reals are a matter of faith with me.

I must admit that I am having difficulty following parts of your argument. My fault I am sure.

I have no difficulty following you that mathematicians have decided that it is useful and possible to extend the notation of exponentiation and radicals beyond their original meanings. That kind of extension has happened many times in mathematics.

Nor would it surprise me to learn that the extension means that the laws of exponents must be re-stated in such a way that they reduce to the familiar laws of exponents when restricted to non-negative reals, but that the familiar laws do not apply generally in the extended realm of discourse. I think you have implied that. Is that correct?

I am guessing that the natural log function has also been extended in conjunction with the extension of exponents. Otherwise it makes no sense to say:

\(\displaystyle x^{(1/a)} = e^{(1/a) * ln(x)}\ if\ a\ is\ an\ integer\ and\ x\ is\ any\ complex\ number\) unless ln(x) has been defined for all x. However, the statement is virtuallly self-evident for positive real x so I can see why that may be an interesting route for extension.

Am I with you so far?

I am further guessing that the extension of the natural log function takes the form of a recursive definition where \(\displaystyle \ln(x) = [\ln(|x|] + i\theta\ for\ all\ x.\)

How am I doing so far? I see what is happening with the positive reals. Theta is zero so the whole thing reduces to the familiar. I see what is happening (without saying that I could justify it or even really understand the rationale) for the negative reals: we are rotating in the complex plane to flip from the positive portion of the real line to the negative portion of the real line.

I also get that you are saying that it is a matter of convention that, when one root out of several is to be identified, the principal complex root is to be identified (although I gather from pka that the convention is not universally accepted). But in the case at hand, we have an equation to be solved. It has four solutions, two real, two complex. (I admit I completely overlooked the two complex solutions and so gave an incomplete answer.)

I say \(\displaystyle x^2 = 4\ entails\ x = \pm \sqrt{4} = \pm 2\)

I simply do not see that \(\displaystyle x^2 = 4\ entails\ x = 2\)

Moreover, if as a matter of convention, a single answer must be given, then you say it should be the principal complex root, which means both 20.032 and 19.968 are incorrect answers.

Or if a real answer is required, why is 20.032 to be preferred over 19.968?

Finally, if by the conventions of notation, \(\displaystyle Solve\ for\ z\ where\ (\dfrac{z}{4} - 5)^{(2/3)} = 0.04\) requires a single solution

how does one ask for all the solutions?

The principal complex root is real if x is non-negative, so 20.032 is the answer. 19.968 is only correct if one defines \(\displaystyle x^{a/b} = \sqrt{x^a}\) when \(\displaystyle x<0\) which can cause "paradoxes" such as the following:

\(\displaystyle -1 = (-1)^{3} = (-1)^{6/2} = \sqrt{(-1)^6} = \sqrt{1} = 1\)

or, if you'd like, for our present situation:

\(\displaystyle -1 = ((-1)^{2/3})^{3/2} = (\sqrt[3]{(-1)^2})^{3/2} = 1\)

There I used the misstep that \(\displaystyle (-1)^{2/3} = 1\).. It is not true.

Mathematicians want everything to make sense for all complex numbers and not just special cases, since they encompass everything that is important and presumably necessary to investigating the real numbers.

If one were to solve this problem for all complex solutions, there would still be only one, and it would be real. Reason being that the right-hand side is real, and if what is in the parentheses on the left hand side is not positive, the entire left-hand side is non-real, so it is an impossibility.

 

[lookagain]

\(\displaystyle -1 = (-1)^{3} = (-1)^{6/2} = \sqrt{(-1)^6} = \sqrt{1} = 1\)

or, if you'd like, for our present situation:

\(\displaystyle -1 = ((-1)^{2/3})^{3/2} = (\sqrt[3]{(-1)^2})^{3/2} = 1\)

There I used the misstep that \(\displaystyle (-1)^{2/3} = 1\).. It is not true.

\(\displaystyle -1 = (-1)^{3} = (-1)^{6/2} = (\sqrt{-1})^6 = (i)^6 = i^2 = -1\)




\(\displaystyle -1 = ((-1)^{3/2})^{2/3} = [(\sqrt{-1})^3]^{2/3} = [(i)^3]^{2/3} = (-i)^{2/3} = (\sqrt[3]{-i})^2 = (i)^2 = -1\)