Mathematical Olympiad | Solve for a+b | Math Olympiad Preparation

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Mathematical Olympiad | Solve for a+b | Math Olympiad Preparation

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[stQZuO]

I think this is a better approach: notice that 9=3^2 = (ab)^2. The first equation can be rewritten as: a^2-b^2 = (ab)^2, we get a quadratic equation of (a/b). Solve this equation and together with ab=3, it's easy to find the value of a and b and consequently a+b.

[victorperesmendosa]

The last thing that I remember is that I was looking for a song....

[georgebliss]

a^2 - b^2 = 9
ab = 3

b = 3/a
a^2 - (3/a)^2 = 9
a^2 - (9/a^2) = 9
Let a^2 = x
x - (9/x) = 9
Multiplying by x
x^2 - 9 = 9x
x^2 - 9x - 9 =0
Solve for x.
Then a = square root of x
Then b = 3/a

[jimlocke]

From the second equation, b = 3/a. Substitute the value of b into the first equation: a² + (3/a)² = 9. Multiply both sides by a² to get a⁴ + 9 = 9a². Rearrange to get a⁴ - 9a² +9 = 0. Let x = a² to make this a quadratic: x² -9x +9 = 0. Solve with quadratic formula x = (9 + √(81 + (4)(1)(9))/2 or x = (9 - √(81 + (4)(1)(9))/2 simplifying to x = (9 + 3√(13))/2 or x = (9 - 3√(13))/2. Replace x by a² and note that the second value of x is negative. Ruling out imaginary numbers for a, x must be positive, a² = (9 + 3√(13))/2. From the first equation, b² = ((9 + 3√(13))/2) - 9 which simplifies to b² = (3√(13)-9)/2. we note that, for ab = 9, a and b must both be positive or both negative. Taking the positive roots, a + b = √((9 + 3√(13))/2) + √((3√(13)-9)/2). Using a scientific calculator, a = 3.147749500 and b = 0.953061862. a + b = 4.100811362. PreMath's solution of a + b = 6 + 3√(13) = 4.100811362 is identical when calculated to the same level of precision. I suspect that my solution can be simplified to be identical to PreMath's solution.

Taking negative values for my a and b produces a solution with the same magnitude but a negative sign, PreMath's other solution.

[danyar]

I followed the same approach, but a^2 + b^2 cannot be equal to -sqrt(117), since it is always positive.

[MrMichelX]

Brilliant ! I like the fact that the explanation is always easy with you, step by step, reminding the rules.

[abeonthehill]

Your explanation is so “ step by step” and i actually follow the procedures !
Thank you for sharing ……Abe ( uk )

[tobiasst]

Great video as always, found it even trickier than most of your other videos 😁
I am wondering: Could you solve a task like this one quickly if you were asked? I think it is anything but straight forward.

[sivanaidoo]

An easier way is to use a right angles triangle with a as hypotenuse and the other 2 sides as b and 3. This uses the given info which is a^2 - b^2 = 9. Use
the second info ab = 3 to get b = 3/a. Now use the right angle triangle to get a^2 - (3/a)^2 = 9. In the calculation use a substitution x = a^2. Now solve a quadratic eqn to get x and then solve for a and b. Now do a + b. Very very simple. Unlike the method used in this video. Thanks. Have a great day.

[francois]

consider the system (1)
a ^ 2 + b ^ 2 = k (where ^ 2 stands for squared)
a ^ 2 - b ^ 2 = 9
from which adding and subtracting the two equations member by member
a ^ 2 = (k + 9) / 2 and b ^ 2 = (k - 9) / 2 (2)
a ^ 2 * b ^ 2 = (ab) ^ 2 = 3 ^ 2 = 9,
applying the (2)
k ^ 2 - 81 = 36 k ^ 2 = 117 = 13 * 9 and k = ± 3 √13
from (a ^ 2 + b ^ 2)> 0, k = + 3√13 follows
a ^ 2 + b ^ 2 = (a + b) ^ 2 -2ab = + 3√13 and
(a + b) ^ 2 = 6 + 3√13 (the solution 6 - 3√13 must be discarded because it is negative in contrast to the first member of equality which is always positive)
a + b = ± √ (6 + 3√13)

[charlesmitchell]

That was a fair amount of work. Good explanation

[vsevolodivanov]

A shorter solution: Substitute x=a+b, y=a-b. We obtain from here that xy=9, x^2-y^2=12. We can remove y from this system to obtain a biquadratic equation for x

[mrshodz]

I have seen some of videos. How do you come up with a strategy for solving these equations? The connection from the start to the end is not obvious. What kind of intuition to you use?

[rcnayak_]

We can still simplify the problem in the following manner. Let x = a² and y = b². So that x - y = a² - b² = 9 ... (1). Again xy = a².b²=(ab) ² = 3² = 9 ...(2) Now using the algebraic equation (x + y )² = (x - y )² + 4xy, we get (x + y )² = 9² + 4. 9 = 117. So that (x + y ) = √117. Here negative sign is not considered as x and y are square terms of a and b . Replacing the values of a and b for x ad y, we get a² + b² = √117. Now (a + b) ² = (a² + b²) + 2ab = √117 + 2. 3 = √117+6. Therefore a + b = +-√ (√117+6), thus our result.

[eben-ezer.hodonou]

Shorter solution : i is the imaginary number
(a + ib)^2 = a^2-b^2+2iab
= 9+6i
Then |a + ib|^2 = sqrt( 81 +36 ) = 3*sqrt(13)

ie a^2 + b^2 = 3*sqrt(13)

a + b = + or - sqrt( a^2 + b^2 + 2ab)
= + or - sqrt(3*sqrt(13) + 6)

[nikitammf]

Let a + b = c. By an express from this equality `a`(b) and squaring both sides we are getting a^2 - b^2 = c^2 - 2ac and b^2 - a^2 = c^2 - 2bc. Hereof c^2 - 9 = 2bc and c^2 + 9 = 2ac, from product of the last two equations are getting c^4 -12 c^2 - 81 = 0. By solved this biquadratic equation we are getting the same solution.

[vijayarya]

b=3/a
a^2-(3/a)^2=9
a^2-9/a^2=9
a^4-9a^2-9=0
a^2=(9+3sqrt13)/2. (a^2 is +ve -sign is Cancel)
Now putting the value of a^2 in Ist eqaution

[awandrew]

I suppose it would be a lot easier if you use the concept a square-b.square= (a+b)(a-b)=9 and it gives a+b =9/a-b, given the ab =3, a+b =9/a-a/3, a+b=27/2a, giving a.square=81/2, giving a=9/root2 , b now=3/a=3xroot2/9=root2/3; therefore.a+b= 9/root2+root2/3, a=b =29xroot2/3

[cxgclhb]

You can go directly from a^4 - 2a^2b^2 + b^4 = 81 to a^4 + 2a^2b^2 + b^4 = 81 + 4a^2b^2. The LHS is (a^2 + b^2)2 and the RHS is 117. So instead of adding 18 twice you just add 36 once. And you don't need to consider a^2 + b^2 = - 3 sqrt(13). You keep that option around far longer than necessary.

[ewerest]

I love it!!! Very similar to "the answer is made by rotating a roulette N^((i*pi*e)^-i) times" Because exist traditional method solving this question