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thedarktiger

Member

Registered: 2014-01-10

Posts: 91

some hard algebra problems

1) Compute the sum

2) Find the ordered quintuplet (a,b,c,d,e) that satisfies the system of equations:

3) Suppose p+q+r = 7 and p^2+q^2+r^2 = 9. Then, what is the average (arithmetic mean) of the three products pq, qr, and rp?

4) Find the largest four-digit value of t such that

is an integer.

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Bob

Administrator

Registered: 2010-06-20

Posts: 10,621

Re: some hard algebra problems

hi thedarktiger

Wow!  I see your algebra has come on a lot. 

1) and 4) are beyond my powers but others will hopefully jump in on those and give you (and me) an idea what to do.

2)  I don't have a super number cruncher for this, so I'd have to slog through elimination.  Do-able but tedious.

3)  Arhh! One I can contribute to.  If you expand (p+q+r)^2 you'll find you can re-arrange this so that pq+qr+rp is the subject and the other side of your equation can be substituted with numbers.  So it's easy from there.

Bob

Might have an idea for 2).  I'll be back if it works. 

LATER EDIT:  Yes it does.

Add all five equations and simplify to get the value of a+b+c+d+e

Now subtract the first from the second, substitute in for  a+b+c+d+e and you'll get the value of e.

Do similar tricks with subtractions of other pairs of equations to get other letters.

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Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

phrontister

Real Member

From: The Land of Tomorrow

Registered: 2009-07-12

Posts: 4,879

Re: some hard algebra problems

Hi thedarktiger,

1) I added up the whole series in Mathematica and got 45, but I can't think of a good way of solving this problem by hand.

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"The good news about computers is that they do what you tell them to do. The bad news is that they do what you tell them to do." - Ted Nelson

Bob

Administrator

Registered: 2010-06-20

Posts: 10,621

Re: some hard algebra problems

1)  Use the difference of two squares thus:

Do that for every term, cancel all but the first and last, and it comes out (as 45, thanks Phro )

4)Let that expression = N

Then

Using the quadratic formula you want b^2 - 4ac to be a perfect square.

So you just need to find the biggest t (under 10000) that gives such a perfect square.

Bob

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Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

thedarktiger

Member

Registered: 2014-01-10

Posts: 91

Re: some hard algebra problems

Thank you guys so much!
These problems were completely beyond me...

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Good. You can read.

thedarktiger

Member

Registered: 2014-01-10

Posts: 91

Re: some hard algebra problems

on problem 3 I took
(p+q+r)^2 = 7^2
and
p^2+q^2+r^2 = 9
and set
(p+q+r)^2-40 = p^2+q^2+r^2
then
(p^2+2 p q+2 p r+q^2+2 q r+r^2)-p^2-q^2-r^2 = 40
and 2 p q+2 p r+2 q r = 40
p q+p r+q r = 20,
but it askes for the average so the answer is 10, right?
well it says that is incorrect

---

Good. You can read.

Bob

Administrator

Registered: 2010-06-20

Posts: 10,621

Re: some hard algebra problems

hi thedarktiger

If pq + qr + rp = 20 (correct so far) you need to divide by three for the average.

Bob

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Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

phrontister

Real Member

From: The Land of Tomorrow

Registered: 2009-07-12

Posts: 4,879

Re: some hard algebra problems

Hi Bob,

Do that for every term, cancel all but the first and last, and it comes out...as 45

I don't understand what you've done there.

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"The good news about computers is that they do what you tell them to do. The bad news is that they do what you tell them to do." - Ted Nelson

Bob

Administrator

Registered: 2010-06-20

Posts: 10,621

Re: some hard algebra problems

hi phro,

Funnily enough it was your answer of 45 that made me think properly about this one.  It's got loads of irrationals from the roots, but somehow they must all cancel out and leave a simple number.  So I started to think seriously about how that might happen.

If you times each term by a fraction thus

then the denominators simplify to just 2.  Thus the roots are moved to the numerators and with a minus sign now instead of a plus.  It's a bit like rationalising a surd.

Now put all over the common denominator of 2, and you get a load of cancelling like this

The only terms that don't cancel are -root(100) at the start and + root(10000) at the end.  As both have integer roots that does it. 

Bob

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Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

phrontister

Real Member

From: The Land of Tomorrow

Registered: 2009-07-12

Posts: 4,879

Re: some hard algebra problems

Hi Bob,

Oh, I see! Excellent deduction!

If I possessed that kind of insight I could throw my M into the bin!

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"The good news about computers is that they do what you tell them to do. The bad news is that they do what you tell them to do." - Ted Nelson

Bob

Administrator

Registered: 2010-06-20

Posts: 10,621

Re: some hard algebra problems

Eeeekkk!  Don't do that.  Without your answer, I wouldn't have thought I could do the problem. 

Teamwork! 

Bob

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Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob