Math Olympiad | A Very Nice Algebra Problem

Math Olympiad | A Very Nice Algebra Problem

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[danieldepaula]

I think an easier way would be just defining c=sqrt(a) and d=sqrt(b). After that, we have a simple system of sum (S) and product (P) that can be solved by the equation x²-Sx+P=0, where the roots are c and d. Then, just find a=c² and b=d².

[alagaika]

Once you have a+b=80, you can substitute a=40-x, b=40+x and enter that into ab=100. Much easier to do without paper, no need for completing the square.

[silver]

1) The equations are obviously symmetrical in a and b, so once you have found the first ordered pair you can write down the second.
2) There was a lot of squaring, so it would be worth checking that the answers found are valid
3) You use sqrt(a)*sqrt(b) = sqrt(a*b) without noting any conditions. OK in this case as both a and b are positive

[Kounomura]

The two-variable equation can be immediately reduced to a one-variable one, due to the symmetries. Let sqrt(a)= 5-x, sqrt(b) = 5+x. Thus (5+x)(5-x)=10 --> 25-x^2 =10 -->
x = sqrt(15) --> a=(5-sqrt(15))^2, b=(5+sqrt(15))^2 or vice versa.

[TheXJ]

Since √ab is non-zero, a is non-zero and I can multiply the first equation by √a
--> (√a)² + √ab = 10√a
--> (√a)² + 10 = 10√a
--> A² + 10 = 10A
then solve A² -10A + 10 = 0

[blue_birb]

I did it orally before opening the video and now I feel better about myself

[arthur_p_dent]

6:27 at this point, if the previous caluculationsa are all correct, then due to the symmetry of the problem, if a=40+10 sqrt(15) then b must be 40-10 sqrt(15) and vice versa.

[ilhmyip]

Элементарно же: из второго уравнения выражаем одну из переменых, подставляем в первое, получаем квадратное уравнение относительно корня, отрицательное значение отбрасываем, положительное возводим в квадрат. Подставляем найденое значение во второе уравнение, находим вторую неизвестную.

[economicsonline]

I watch these videos on double time. Too many easy steps drawn out separately and needlessly. Don't have to show subtracting in two steps. Just do it.

[lakshmikutty]

√a+√b=10 √ab=10
a+b+2√ab=100 a+b=80 a+b==100-20=80
(a+b)² -4ab=(a-b) ² =80²-400
( 80+20) (80-20) =6000
6000=2×3×5×2×10×10
a-b=20√15
a+b=80
2a=80+20√15=20(4+√15)
a=10(4+√15)
b=80-10(4+√15)

[starpawsy]

I would square both equations, then substitute the second into the first. Fairly straightforward from then on.

[yifengxiao]

let √a and √b be two roots of a quadratic equation x² -10x+10=0, and solve it. This idea is applicable to any case with the form of sum and product of two variables. For example, you may solve x³ +y³=5, x³y³=6, or a ∛a + ∛b =6, ∛ab =8 in the same way. just as danieldepaula says.

[lightning]

a = 40+10(sqrt(15))
b = 40-10(sqrt(15))
Square both sides, solve for (a+b) and (ab), use Quad. formula to get answer.

[mmggxhz]

Можно решить в 5 строчек если предствить корни из a и b как корни квадратного уравнения x^2 - 10x +10=0

[youandi]

a + b + 2 akar ab = 100
a + b + 20 = 100
a + b = 80

(akar ab)² = 10²
ab = 100

(a + b)² = a² + b² + 2ab
640 = a² + b² + 200
a² + b² = 440

[gizlnvt]

let sqrt(a) = x, sqrt(b) = y
then x + y = 10, xy = 10 -> (x, y) = (5 + sqrt(15), 5 - sqrt(15)) or reversed.
therefore, (a, b) = (sqrt(5 + sqrt(15)), sqrt(5 - sqrt(15)))

that's the easy way.

[davidbrown]

Thanks for there challenge. Nailed it in one.

[theeternalswrd]

If the problem is asking for real solutions only, you need to show that 10*sqrt(15) < 40 to prove that 40 - 10*sqrt(15) is positive. To do this, divide both sides by 10 so that sqrt(15) < 4 then square both sides 15 < 16, which is true.

[subbaraob]

When we know a+b, అండ్ ab
Wecan find a-b
(a+b)^2- 4ab

[gheffz]

What's this one? Sqrt(ab) = 10 & Sqrt(a) + Sqrt(b) = 7