quadratic

cooljackiec · 2014-03-12 12:39:48

If (x - a)(x - 1) + 1 = (x + b)(x + c) is true for all x and if a, b, and c are integers, find the sum of all possible values of a.

Bob · 2014-03-12 21:46:07

hi cooljackiec

Had several goes at this, but kept making silly arithmetical errors. This version checks out properly.

If this is true for all x, choose x = 0 and we get

and then (1) becomes

So two equations and three 'unknowns'.

Adding gives

and substituting back

This didn't look like much of a restriction on a so I tried graphing it (using y for a and x for b). That graph is shown below.

You can see that other than b ≠ 1(and hence a has no values in the range [-5,-2] there is no restriction on a, so I cannot see how you can add them up and get a sensible answer.

???

Hold a moment. Choosing an integer for b, doesn't necessarily give an integer for a. Just trying to modify the answer to allow for that.

Now I'm only getting b = c = 0 gives a = -1 AND b = -2 c = -2 gives a =3

That's more like it. But are there any more solutions?

No that's the lot. The denominator (1+b) has to be 1 or -1 in order for it to divide b^2 exactly.

Hence values of a are -1 and 3 so their sum is 2

Check.

a = 3 gives

and a = -1 gives

Bob

Nehushtan · 2014-03-12 21:53:54

You made a mistake there should be a minus sign in front of the x on the RHS of your equation (1).

Bob · 2014-03-13 00:13:57

hi Nehushtan,

I found two errors with my first post. Post 2 was the corrected version. Have you found another error?

Bob

Nehushtan · 2014-03-13 00:34:06

Hmm, it seems I was looking at your original post before you corrected your errors. Sorry about that. False alarm.

Bob · 2014-03-13 00:41:05

No need to apologise.

I don't think my brain was fully awake when I started the problem. I had about 5 edits before I got to any answers at all.

I hope it's all ok now

Bob

gourish · 2014-03-15 05:24:12

hi bob bundy i really struggle at such questions whenever i see "find the possible number of values of a for blah blah equation having a as coefficient" can you suggest me any way of tackling such questions btw i did not understand your answer from the part after the graph like how did you restrict the value of a?

Bob · 2014-03-15 06:16:21

hi gourish,

To start with I thought there were an infinite number of solutions for 'a' and my first attempt at the post said that. Sometimes after I've made a post I realise I've missed something and I have to log back in to correct it. In this case I had lots of attempts before I got an answer I was happy with.

I made a spreadsheet with columns for values of a, b and c. I put the formula for a in terms of b, and similarly the formula for c in terms of b. That way I only had to try different values for b to see what happens.

I set up the b column with -10, -9, -8, ......., 0, 1, 2, 3, ..........20 and used the formulas to generate the values for a and c. It was immediately obvious that only two values for b gave integers for a and c, so that set me on the right track. Then I had to think up a reason why no other values of b would give integers.

I realised that large values of b^2 would never be divisible by (b+1) so I knew then I could justify my statement that I had found all the solutions.

So, as you can see now I've revealed my methods, there was no clever insight, just lots of stumbling around in the dark, trying to find the light switch.

I'm having a similar problem with thedarktiger's question involving triangles. I've made a diagram and tried varying the initial triangle. I've measured lots of angles and checked for parallels and points on circles. So far no answer.

bobbym has some useful advice about problems. It used to be in his signature but he has changed that recently. I may have the wording slightly wrong but the gist is this: "Check it out with numbers." As bobbym reads everything, he will, no doubt, give us the correct quotation. thanks bobbym in anticipation.

Bob

gourish · 2014-03-15 06:37:19

thanks bob... well if i want to plot graphs of a function (generally complex and mind boggling) what is the best way to do it on my own i wanna skip the machine part of it

ShivamS · 2014-03-15 06:39:00

Drawing such graphs is pretty hard. If I were you, I'd just use Wolfram.

Bob · 2014-03-15 06:40:58

Here's some homework for you:

http://www.mathisfunforum.com/viewtopic.php?id=15139

Good luck

Bob

bobbym · 2014-03-15 08:05:10

Bob Bundy wrote:

bobbym has some useful advice about problems. It used to be in his signature but he has changed that recently. I may have the wording slightly wrong but the gist is this: "Check it out with numbers."

He might have had a quote like that or even said something along those lines. Even a blind squirrel finds an acorn now and then.

Bob · 2014-03-15 08:30:28

Thanks bobbym. You had it as part of your sig. It sounded much better then so please try to remember for me. Thanks.

Bob

bobbym · 2014-03-15 10:19:13

I remember it vaguely, I got it out of a book but do not remember where.

Math Is Fun Forum

#1 2014-03-12 12:39:48

quadratic

#2 2014-03-12 21:46:07

Re: quadratic

#3 2014-03-12 21:53:54

Re: quadratic

#4 2014-03-13 00:13:57

Re: quadratic

#5 2014-03-13 00:34:06

Re: quadratic

#6 2014-03-13 00:41:05

Re: quadratic

#7 2014-03-15 05:24:12

Re: quadratic

#8 2014-03-15 06:16:21

Re: quadratic

#9 2014-03-15 06:37:19

Re: quadratic

#10 2014-03-15 06:39:00

Re: quadratic

#11 2014-03-15 06:40:58

Re: quadratic

#12 2014-03-15 08:05:10

Re: quadratic

#13 2014-03-15 08:30:28

Re: quadratic

#14 2014-03-15 10:19:13

Re: quadratic

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