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A funny algebra puzzle!

a and b are real numbers

We have the following expressions:

a² - b² = 9

a.b = 3

Question: what are the value(s) for a and b ?

Good luck! And try to be among the first ones!

Hide replies until user replies.
[edited by: emassa at 2:37 PM (GMT -5) on 20 Feb 2023]
• 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...