# Need help real bad math

the following two questions are in math..and i have no clue how to even start..
i got them in a test..
i would appreciate it if u could tell me what i should do , or what concept is involved...

1) if a,b,c belong to the set of all real no.s and a!=b(not equal to)
then

a)a^2+b^2+ab=3c^2 for some c in (a,b)
b)a^2+b^2+ab=3c^2 for not even one c in (a,b)
c)a^2+b^2-ab=3c^2 for some c in (a,b)

2)the no. of non-negative integral solutions for the eqn.
2x+y+z=20 are
a)132
b)121
c)144
d)none of the above

## Answers and Replies

bogdan
1) Take a=0,b=1
a^2+b^2+ab=1;3*c^2=1 => c is in (a,b);
Correct answer : a)

2) x can be 0,1,2,3,...10

y+z can be 20,18,16,...4,2,0;
y can take 21,19,17,...5,3,1 values for every y+z;
So 21+19+17+...+3+1=121...
Correct answer : b)

[?]

Homework Helper
You have to be a little more careful, Bogdan. Showing that there exists such a c for a=0 b= 1 doesn't show that there is such a c for any a,b.

(a+b)2= a2+ 2ab+ b2 so

a2+ b2+ ab= (a+b)2- ab

If (a+b)2> ab (a=1, b=0 for example then the right side is positive and there exist c such that 3c2 is equal to it.

If (a+b)2< ab (example, a= 1, b= -1) then the right hand side is negative and there is no such c (3c2 can't be negative).

The correct answer is b.

STAii
Originally posted by bogdan
2) x can be 0,1,2,3,...10

y+z can be 20,18,16,...4,2,0;
y can take 21,19,17,...5,3,1 values for every y+z;
So 21+19+17+...+3+1=121...
Correct answer : b)
Sorry but ... how did u reach this conclusion ?
I haven't reached any answer yet, but i have an approach.
x => 0
y => 0
z => 0
Let's first look at z.
Rearrange the original equation.
2x+y+z=20
z=20-2x-y
Therefore as x and y get smaller, z will get bigger.
So a the lower bound of x and y, you will have the upper bound of z.
The lower bound of x is 0, the lower bound of y is 0 too.
So the upper bound of z is 20-(2*0)-(0) = 20
So now we have a smaller range of z.
0 <= z <= 20
Do the same for x and y, you will reah the following conclusions.
0 <= x <= 10
0 <= y <= 20
And ... i am still thinking of remaining (here is a clue, for each value of x, a certain value of y+z will be determined, just find the number of posibilities for y+z on each value of x (of course not by trial and error )).
I will work on it and reply ASAP.

S.P.P
121

i get 121 for the last question.

bogdan
Maybe I'm wrong...
But b) can't be (busta rhymes...) the correct answer...
I demonstrated that for a=0, b=1,c=sqrt(1/3)...so there is a c in (a,b)...there is an example...mine...
I think the problem is if the answer is a or c...because a=0 b=1 makes a and c correct...take a=2,b=3...and you'll probably find out that a is the correct answer...
Correct me if I'm wrong...
mercury, do you have the correct answers ?

STAii
I was finally able to continue my method (which i started in the last post) to find the answer of the question (of the second question).
The answer is 121.
Let me show you how ...
From the last post, i showed that
0 <= x <= 10
now suppose you wanted to solve the question by trial (try each value of x, and find the possible values of y and z at each value of x), you will get something like this.
(please note that if you are given a value for x, and a value of y, there will be only a single value for z that satisfies the equation, therefore it is not needed to list the values of z)
> x=10 (y+z=0)
>> only one posibility, which is that y = 0
> x=9 (y+z=2)
>> posibility one : y=0
>> posibility two : y=1
>> posibility three : y=2
> x=8 (y+z=4)
>> posibility one : y=0
>> posibility two : y=1
>> posibility three : y=2
>> posibility four : y=3
>> posibility five : y=4
... so on ...
so far you have the following answers
(x,y,z)

(10,0,0)

(9,0,2)
(9,1,1)
(9,2,0)

(8,0,4)
(8,1,3)
(8,2,2)
(8,3,1)
(8,4,0)

... so on ...
Now you may see that the number of possibilities of y and z for each value of x can be expressed as
P=21-2x
(this is very logical **)
So the number of answers is
[sum](from x=0 to x=10)(21-2x)
=21*(10+1)[sum](from x=0 to x=10)(-2x)
=21*11 -2 * [sum](from x=0 to x=10) (x)
=21*11 -2*(10+1)*(10/2)
=231-2*(11)*(5)
=231-110
=121

Staff Emeritus
Gold Member
If (a+b)2 < ab (example, a= 1, b= -1)

Slight problem with this statement... the left hand side of this inequality is 0 while the right hand side is negative. It doesn't hold!

a) is correct for problem 1:
(without loss of generality, let a < b)

case 1: 0 <= a < b

notice that 3a^2 < a^2 + b^2 + ab < 3b^2, because:

a^2 = a^2 < b^2
a^2 < b^2 = b^2
a^2 < ab < b^2

Since 3c^2 is a continuous function, the intermediate value theorem says that there is a c such that:

3c^2 = a^2 + b^2 + ab

case 2: a < b <= 0

This time, 3b^2 < a^2 + b^2 + ab < 3a^2, and the same argument holds.

case 3: a < 0 < b

3a: |a| < |b|

3*0^2 < a^2 + b^2 + ab < 3b^2

Therefore there is a c in (0, b) for which equality is satisfied, and thus c is in (a, b)

3b: |b| < |a|

3*0^2 < a^2 + b^2 + ab < 3a^2

Therefore there is a c in (a, 0), and thus c is in (a, b)

3c: |a| = |b|

a^2 + b^2 + ab = b^2 + b^2 - b^2 = b^2

Setting c = b / sqrt(3) yields equality, and c is in (a, b)

Hurkyl

bogdan
Thank you Hurkyl for the time spent proving rigorously the first problem...as I said...it was a)

hey people...thanks loads for answering my question and taking a lot of pains to explain it in detail...i have learned a lot of new things..
i suppose that's what this forum is for ...?

anyway i appreciate it..

cheers and peace

mercury