Thread: Need to clarify a quadratic equation (simple)

Is it true that in this equation:



To make the fraction a whole number (x being a multiple of 99) it is not possible to make the (a-b) and (a+b) make a multiple of 99, with a and b being single digit numbers only.

I don't think so myself. The only possible situation would be used two digit numbers e.g. a as 10 and b as 1. However this would defy the question.

Why would you want a- b or a+ b to be a multiple of 99? The left side is not likely to be!

The right side is , a difference of squares.
Neither 1 nor 2 is a difference of squares but 3 is: you could have x= 3(99)= 297, a= 2, b= 1. 4 is not a difference of squares but 5 is: you could have x= 5(99)= 495, a= 3, b= 2.

Am I misunderstanding your question?

Originally Posted by Mukilab

Is it true that in this equation:



To make the fraction a whole number (x being a multiple of 99) it is not possible to make the (a-b) and (a+b) make a multiple of 99, with a and b being single digit numbers only.

I don't think so myself. The only possible situation would be used two digit numbers e.g. a as 10 and b as 1. However this would defy the question.

1. I assume that . Thus .

2. If then

3. Therefore you'll get these single digit results:



4. If then

5. Therefore you'll get these single digit results:



6. Use the same way to dtermine the values of a and b if




7. If then you have to use at least two-digit-numbers.