# Solving Real Number Equations and Counting Integral Solutions: Help Needed!

• mercury
In summary, the first problem involves finding the correct answer, which is a), by considering different cases and using the intermediate value theorem. The second problem can be solved by finding the number of possible solutions for x, y, and z, which is 121.

#### mercury

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

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

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...

[?]

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).

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...
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.

121

i get 121 for the last question.

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 ?

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).
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

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

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

## 1. What are real number equations?

Real number equations are mathematical expressions that contain real numbers, which are numbers that can be expressed on a number line and include both positive and negative numbers, fractions, and decimals.

## 2. How do you solve real number equations?

To solve a real number equation, you must isolate the variable on one side of the equation and perform inverse operations to cancel out any constants or coefficients. The final value of the variable will be the solution to the equation.

## 3. What are integral solutions?

Integral solutions are solutions to an equation that are whole numbers, both positive and negative. They are also known as integer solutions.

## 4. How do you count the integral solutions to an equation?

The number of integral solutions to an equation can be counted by looking at the highest degree of the variable in the equation. The number of solutions will be equal to the degree of the variable in the equation.

## 5. Can you provide an example of solving a real number equation and counting integral solutions?

For example, if we have the equation 3x + 5 = 14, we can solve for x by subtracting 5 from both sides and then dividing by 3, giving us x = 3. This is an integral solution since it is a whole number. The equation 2x^2 - 8 = 0 has two integral solutions, x = 2 and x = -2.