A funny algebra puzzle!

From a.b=3: a=3/b

Substitute a=3/b to a^2-b^2=9

(3/b)^2 - b^2 = 9

9/b^2 - b^2 = 9

Multiply both sides by b^2

9 - b^4 = 9b^2

b^4 + 9b^2 - 9 = 0

Solving for the roots will obtain the following:

b1 = √[(-9+3√13)/2]

b2 = -√[(-9+3√13)/2]

b3 = √[(-9-3√13)/2]

b4 = -√[(-9-3√13)/2]

Since the square root of a negative number is imaginary, b3 and b4 are not answers

Substitute the values of b to obtain a:

a1 = 3/√[(-9+3√13)/2]

a2 = -3/√[(-9+3√13)/2]

The values for a and b are the following:

a1 = 3/√[(-9+3√13)/2]

a2 = -3/√[(-9+3√13)/2]

b1 = √[(-9+3√13)/2]

b2 = -√[(-9+3√13)/2]

a² - b² = (a+b)(a-b) = 9 =3x3

(a+b) = 3 
(a-b) = 3

2a = 6

a=3

b=0

cannot be true as b=3/a = 1

Starting with:

a² – b² = 9

a.b = 3

From second equation:

b = 3/a

b² = 9/a²

Substitute into first equation:

a² – 9/a² = 9

Subtract 9 from both sides:

a² – 9/a² – 9 = 0

Multiply top and bottom by a²

a⁴ – 9a² – 9 = 0

Let a² = x

x² – 9x – 9 = 0

General quadratic equation ax² + bx + c = 0

a = 1; b = – 9; c = – 9

Solved by formula x = [–b ± √ (b² – 4ac)] / 2a

x = (9 ± √(81+36)) / 2 = (9 ± √(117)) / 2

Take out a factor of 3 from top

x = 3(3 + √13) / 2 or x = 3(3 – √13) / 2

So

x = 9.908…

x = –0.908

As x = a²

a = √x

so a = 3.148

b = 3 / a

b = 0.953

Thank you Gaetano! As recommended by our Engineer Zone administrators, the final answers will be confirmed later (just before the new quiz launch; this always in order to not influence or perturb our other participants who are still trying to answer...

Many thanks Fons! It's indeed a little bit more complex with the dot line in a.b not being a simple decimal point...As stated for Gaetano, the final answer will be confirmed in a few days...

Thanks a lot Sona for your prompt reply! Your answer will be confirmed in a few days! Stay tuned!

Thanks Joachim for your prompt reply! You are right with a=5, b=4 for the first equation (a²-b²=9). But they are not OK for the second equation a*b=3 since 4*5= 20 !

May be you want to retry?

Thanks Mark for your extensive math development. As stated earlier, all the answers will be confirmed (or not) in a few days when the new quiz will be launched. This, mainly to let everybody the chance to try to solve the puzzle....

Hello Rajesh! Happy to hear from you again! Thanks to have proposed your answers to this math puzzle (not so simple at the end!). However, what I can say here is : there well real solutions for these equations! Your development is correct but you started with an incomplete assumption already from the beginning. In effect a²-b² can be split as (a-b)*(a+b). BUT then, it is very limitative to just say a+b = 3 and a-b=3. This was because you have assumed only integer values in the equation. There are infinite number of real values a and b satisfying a²-b²=9. May be you want to retry?

Thanks Martin for your so detailed development:! As stated previously: "rendez-vous" in few days to see if the anwers are confirmed!