what's wrong here? I get "no real roots" for sqrt{2z} - sqrt{z + 7} = -1

[allegansveritatem]

Here is the equation to be solved for x--it is actually z but I use x in my handwritten working out (attached) to avoid confusion of z with 2:

\(\displaystyle \mbox{41.}\qquad \sqrt{\strut 2z\,}\, -\, \sqrt{\strut z\, +\, 7\,}\, =\, -1\)

Here is what I did--and this is only one version because I turned this every way but loose trying to figure it out:

. . . . .\(\displaystyle \sqrt{\strut 2z\,}\, -\, \sqrt{\strut z\, +\, 7}\, =\, -1\)

. . . . .\(\displaystyle -\sqrt{\strut z\, +\, 7\,}\, =\, -\sqrt{\strut 2z\,}\, -\, 1\)

. . . . .\(\displaystyle \left(-\sqrt{\strut z\, +\, 7\,}\right)^2\, =\, \left(-\sqrt{\strut 2z\,}\, -\, 1\right)^2\)

. . . . .\(\displaystyle z\, +\, 7\, =\, 2z\, +\, \sqrt{\strut 2z\,}\, +\, \sqrt{\strut 2z\,}\, +\, 1\)

. . . . .\(\displaystyle z\, +\, 7\, =\, 2z\, +\, 2\, \sqrt{\strut 2z\,}\, +\, 1\)

. . . . .\(\displaystyle z\, +\, 6\, =\, 2z\, +\, 2\, \sqrt{\strut 2z\,}\)

. . . . .\(\displaystyle -z\, +\, 6\, =\, 2\, \sqrt{\strut2z\,}\)

. . . . .\(\displaystyle (-z)^2\, +\, (6)^2\, =\, (2)^2\, \left(\sqrt{\strut2z\,}\right)^2\)

. . . . .\(\displaystyle z^2\, +\, 36\, =\, 4\, \times\, 2z\)

. . . . .\(\displaystyle z^2\, -\, 8z\, +\, 36\, =\, 0\)

. . . . .no real roots??

I used a calculator to check this and this is what it said: "No real roots."

Here is the book answer: {2}

It must be something fairly simple I am missing here because this is a simple problem. But, apparently I am simpler than it is. What is it?

 

[Dr.Peterson]

Here is the equation to be solved for x--it is actually z but I use x in my handwritten working out (attached) to avoid confusion of z with 2:

\(\displaystyle \mbox{41.}\qquad \sqrt{\strut 2z\,}\, -\, \sqrt{\strut z\, +\, 7\,}\, =\, -1\)

Here is what I did--and this is only one version because I turned this every way but loose trying to figure it out:

. . . . .\(\displaystyle \sqrt{\strut 2z\,}\, -\, \sqrt{\strut z\, +\, 7}\, =\, -1\)

. . . . .\(\displaystyle -\sqrt{\strut z\, +\, 7\,}\, =\, -\sqrt{\strut 2z\,}\, -\, 1\)

. . . . .\(\displaystyle \left(-\sqrt{\strut z\, +\, 7\,}\right)^2\, =\, \left(-\sqrt{\strut 2z\,}\, -\, 1\right)^2\)

. . . . .\(\displaystyle z\, +\, 7\, =\, 2z\, +\, \sqrt{\strut 2z\,}\, +\, \sqrt{\strut 2z\,}\, +\, 1\)

. . . . .\(\displaystyle z\, +\, 7\, =\, 2z\, +\, 2\, \sqrt{\strut 2z\,}\, +\, 1\)

. . . . .\(\displaystyle z\, +\, 6\, =\, 2z\, +\, 2\, \sqrt{\strut 2z\,}\)

. . . . .\(\displaystyle -z\, +\, 6\, =\, 2\, \sqrt{\strut2z\,}\)

. . . . .\(\displaystyle (-z)^2\, +\, (6)^2\, =\, (2)^2\, \left(\sqrt{\strut2z\,}\right)^2\)

. . . . .\(\displaystyle z^2\, +\, 36\, =\, 4\, \times\, 2z\)

. . . . .\(\displaystyle z^2\, -\, 8z\, +\, 36\, =\, 0\)

. . . . .no real roots??

I used a calculator to check this and this is what it said: "No real roots."

Here is the book answer: {2}

It must be something fairly simple I am missing here because this is a simple problem. But, apparently I am simpler than it is. What is it?

When you squared (-x + 6), you forgot to either use the formula for the square of a binomial, or "FOIL" (-x + 6)^2. The middle term is missing.

But there are a lot of other mistakes I might have expected someone to make, and you got past all those potholes, so congratulations on all the parts you did right! There's some really good work here.

 

[allegansveritatem]

When you squared (-x + 6), you forgot to either use the formula for the square of a binomial, or "FOIL" (-x + 6)^2. The middle term is missing.

But there are a lot of other mistakes I might have expected someone to make, and you got past all those potholes, so congratulations on all the parts you did right! There's some really good work here.

Thanks for pointing this out. I see it but I can't work this out tonight but tomorrow I will have another go at this problem with what you say here in mind. I thnk I see light at the end of the tunnel. Thanks again.

 

[Steven G]

Here is the equation to be solved for x--it is actually z but I use x in my handwritten working out (attached) to avoid confusion of z with 2:

\(\displaystyle \mbox{41.}\qquad \sqrt{\strut 2z\,}\, -\, \sqrt{\strut z\, +\, 7\,}\, =\, -1\)

Here is what I did--and this is only one version because I turned this every way but loose trying to figure it out:

. . . . .\(\displaystyle \sqrt{\strut 2z\,}\, -\, \sqrt{\strut z\, +\, 7}\, =\, -1\)

. . . . .\(\displaystyle -\sqrt{\strut z\, +\, 7\,}\, =\, -\sqrt{\strut 2z\,}\, -\, 1\)

. . . . .\(\displaystyle \left(-\sqrt{\strut z\, +\, 7\,}\right)^2\, =\, \left(-\sqrt{\strut 2z\,}\, -\, 1\right)^2\)

. . . . .\(\displaystyle z\, +\, 7\, =\, 2z\, +\, \sqrt{\strut 2z\,}\, +\, \sqrt{\strut 2z\,}\, +\, 1\)

. . . . .\(\displaystyle z\, +\, 7\, =\, 2z\, +\, 2\, \sqrt{\strut 2z\,}\, +\, 1\)

. . . . .\(\displaystyle z\, +\, 6\, =\, 2z\, +\, 2\, \sqrt{\strut 2z\,}\)

. . . . .\(\displaystyle -z\, +\, 6\, =\, 2\, \sqrt{\strut2z\,}\)

. . . . .\(\displaystyle (-z)^2\, +\, (6)^2\, =\, (2)^2\, \left(\sqrt{\strut2z\,}\right)^2\)

. . . . .\(\displaystyle z^2\, +\, 36\, =\, 4\, \times\, 2z\)

. . . . .\(\displaystyle z^2\, -\, 8z\, +\, 36\, =\, 0\)

. . . . .no real roots??

I used a calculator to check this and this is what it said: "No real roots."

Here is the book answer: {2}

It must be something fairly simple I am missing here because this is a simple problem. But, apparently I am simpler than it is. What is it?

I saw that 2 is an answer by simply looking at the equation. The problem with my method is I do not see another answer that might (or might not) work.

 

[allegansveritatem]

I saw that 2 is an answer by simply looking at the equation. The problem with my method is I do not see another answer that might (or might not) work.

Right. It is easy to eyeball it that way...but that ain't algebra.

 

[allegansveritatem]

So, I went at this again this morning and finally got it to behave. I was just about to post photographic proof but can't find the photo! So you will have to take it on faith: I got it and it was a glorious moment. Thanks again to Dr Peterson for the pointer.

 

[Steven G]

Right. It is easy to eyeball it that way...but that ain't algebra.

I have to agree that what I did was not algebra. It was thinking and that, in my opinion, is more important than algebra. I always spend a couple of seconds to see if there is an easy/obvious answer before I do mechanical thing to arrive at the answer.

 

[Deleted member 4993]

So, I went at this again this morning and finally got it to behave. I was just about to post photographic proof but can't find the photo! So you will have to take it on faith: I got it and it was a glorious moment. Thanks again to Dr Peterson for the pointer.

You should have obtained two solutions for z. Those are:

z = 2 and

z = 18

How come that answer is not reported (neither by the book nor by you)?

 

[Dr.Peterson]

So, I went at this again this morning and finally got it to behave. I was just about to post photographic proof but can't find the photo! So you will have to take it on faith: I got it and it was a glorious moment. Thanks again to Dr Peterson for the pointer.

Perhaps you should show your work, to convince everyone that you did complete it correctly (including checking for an extraneous root), and to help anyone else who might be reading along and need that last bit.

 

[topsquark]

You should have obtained two solutions for z. Those are:

z = 2 and

z = 18

How come that answer is not reported (neither by the book nor by you)?

z = 18 is a spurious solution. It is a solution to
\(\displaystyle \left ( \sqrt{2z} - \sqrt{z + 7} \right ) ^2 = (-1)^2\), which is not a solution to the original equation. Though I grant that this "solution" probably would have been noticed by the OP, but not dealt with. (No offense intended allegansventatem.)

-Dan