am I right here?

[allegansveritatem]

Here is the problem:

Here is my solution (please note that I have changed the z to x so that the z would not be mistaken for a 2):



The reason I am hesitant to accept this as having been worked out or developed as far as it possibly could be is my familiarity with the author's way of pairing an odd numbered problem with the following even numbered problem. The even problem is almost always completely the same as the previous problem except the constants are different. If the odd numbered problem produced a quadratic that could be factored then the even problem will nearly always do the same. Now, the problem previous to this one, number 23, factored nicely. Note also that all the answers to the odd problems are given at the back of the book. With the even numbers you are restricted to your own devices to check your work. The problems with this system are these: 1) When the student works out a problem and finds from the key that it is wrong, he is able to go back and work through the problem again with some idea in mind as to where he may have gone wrong. 2) When he, with great effort, works out a problem but is not certain of it and has no key for it, he is as likely as not to just accept his results and move on. After all, it takes as long to check some problems as it does to work them out. I'm trying to say that it's nice to have some incentive to go back and hack though a knotty problem again, the incentive being that you know: 1) that your first try was wrong and 2) you know what the accepted solution is supposed to look like.

 

[sinx]

Here is the problem:View attachment 10453

Here is my solution (please note that I have changed the z to x so that the z would not be mistaken for a 2):

View attachment 10454

.

I got the same as you,
and could not factor it either.
3z²-12z+8=0

quadratic formula
z=[-b+-sqrt(b²-4ac)]/2a

 

[allegansveritatem]

I got the same as you,
and could not factor it either.
3z²-12z+8=0

quadratic formula
z=[-b+-sqrt(b²-4ac)]/2a

I thought of that...but we have not yet come to that formula in the text. So..I guess the author threw a curve ball here and, well, what's the good of being an author if you can't do that every so often?

Thanks for checking it out.

 

[Denis]

(x - 2 - 2/√3)(x - 2 + 2/√3) = 0 :cool:

 

[Dr.Peterson]

Here is my solution (please note that I have changed the z to x so that the z would not be mistaken for a 2):

Here's a thought: It's more likely that the author (or printer) made a mistake than that this inappropriate problem would be deliberate. What if someone along the line mistook a z for a 2 ...

Try solving z/(z-2) + 4/z = 3.