a²(1-a)=5*5*6. So I expected that 1-a=6 and a²=25, then found a=-5. The rest is just a simple factorization.

But what if x is not a whole number? Could there be a solution if 0 < x < 1?

a,b are the roots of the quadratic equation x²-9x+81=0. Solve the equation and get the values of a and b.

動画見たら14分、コメント欄見たら2秒で解決。便利な世の中になった笑

The 2 you write is similar to 3. I tend to look at the screen without listening to the sound, I get confused between 2 and 3. The horizontal bar of 2 should be written straight.

Clearly, there is no more than one Real solution. (a+b)^2 = a^2 + b^2 + 2ab = 81 if ab = 81, then 2ab = 162. Subbing that in: a^2 + b^2 + 162 = 81 a^2 + b^2 = -81 In the real world, adding two squares always results in a positive sum. Therefore, at least one (if not both) of a and b are complex.

a²-a³=150; solution; a²-a³-150=0; a²-b³-125-25=0; a²-a³-5³-5²=0; a²-5²-a³-5³=0; (a-5)(a+5)-[(a-5)(a²+5a+25)]=0; (a-5)[a+5-(a²+5a+25)]=0; (a-5)[a+5-a²-5a-25]=0; a-5=0; a=5; -a²-4a-20=0;(-1); a²+4a+20=0; S={5}.

Решила устно : - 5

x = log (24) /.log (4) = 3/2 + log (3)/(2 log (2))

😊

Такие "сложные" задачи решаются устно😂, пока Европа расписывает микрошаги😂

a=-5

Let _(t - √x)(t - √y) = 0_ ... ① ⇒ _t² - (√x + √y)t + √(xy) = 0_ ⇒ _t² - 10t + 10 = 0_ ⇒ _t = 5 ± √15_ ⇒ _( t - (5 + √15) )( t - (5 - √15) ) = 0_ ∴ by comparison with ①: _√x = 5 + √15, √y = 5 - √15_ up to symmetry ⇒ *_x = 40 + 10√15, y = 40 - 10√15_*

I did it a different and (unfortunately) longer way. sqrt(x) + sqrt(y) = 10 Since sqrt(x) >= 0, then x >= 0 Since sqrt(y) >= 0, then y >= 0 sqrt(xy) = 10 Since sqrt(xy) > 0, then x > 0 and y > 0 [sqrt(x) + sqrt(y)]^2 = 10^2 [sqrt(x)]^2 + [sqrt(y)]^2 + 2*sqrt(x)*sqrt(y) = 100 x + y + 2*sqrt(xy) = 100 x + y + 2*sqrt(xy) - x - y = 100 - x - y 2*sqrt(xy) = 100 - x - y 2*sqrt(xy)/2 = [100 - x - y]/2 sqrt(xy) = [100 - x - y]/2 [100 - x - y]/2 = sqrt(xy) Since sqrt(xy) = 10, then [100 - x - y]/2 = 10 [100 - x - y]/2*2 = 10*2 100 - x - y = 20 -[100 - x - y] = -20 x + y - 100 = -20 x + y - 100 - y + 100 = -20 - y + 100 x = 80 - y sqrt(x) + sqrt(y) = 10 sqrt(x) + sqrt(y) - sqrt(y) = 10 - sqrt(y) sqrt(x) = 10 - sqrt(y) [sqrt(x)]^2 = [10 - sqrt(y)]^2 x = 10^2 + [-sqrt(y)]^2 + 2*10*[-sqrt(y)] x = 100 + y - 20*sqrt(y) 100 + y - 20*sqrt(y) = x Since x = 80 - y, then 100 + y - 20*sqrt(y) = 80 - y 100 + y - 20*sqrt(y) - 80 + y = 80 - y - 80 + y 20 + 2y - 20*sqrt(y) = 0 2y - 20*sqrt(y) + 20 = 0 [2y - 20*sqrt(y) + 20]/2 = 0 / 2 y - 10*sqrt(y) + 10 = 0 Let u = sqrt(y), then y - 10*sqrt(y) + 10 = u^2 - 10u + 10 = 0 u = [-(-10) +/- sqrt([-10]^2 - 4*1*10)] / [2*1] u = [10 +/- sqrt(100 - 40)] / 2 u = [10 +/- sqrt(60)] / 2 u = [10 +/- sqrt(4*15)] / 2 u = [10 +/- 2*sqrt(15)] / 2 u = 5 +/- sqrt(15) u^2 = [5 +/- sqrt(15)]^2 u^2 = 5^2 + [+/- sqrt(15)]^2 + 2*5*[+/- sqrt(15)] u^2 = 25 + 15 +/- 10*sqrt(15) u^2 = 40 +/- 10*sqrt(15) Since u = sqrt(y), then u^2 = [sqrt(y)]^2 = 40 +/- 10*sqrt(15) y = 40 +/- 10*sqrt(15) y1 = 40 + 10*sqrt(15) and y2 = 40 - 10*sqrt(15) y1 > 0 and y2 > 0 Since x = 80 - y, then x1 = 80 - y1 and x2 = 80 - y2 x1 = 80 - [40 + 10*sqrt(15)] and x2 = 80 - [40 - 10*sqrt(15)] x1 = 80 - 40 - 10*sqrt(15) and x2 = 80 - 40 + 10*sqrt(15) x1 = 40 - 10*sqrt(15) and x2 = 40 + 10*sqrt(15) x1 > 0 and x2 > 0 { (x1, y1), (x2, y2) } = { (40 - 10*sqrt[15], 40 + 10*sqrt[15]), (40 + 10*sqrt[15], 40 - 10*sqrt[15]) }

Since both equations are symmetrical, the two "different" solutions are just notational variants

Why bother finding the complex roots exact values if they are to be discarded? A lot of wasted time. In addition, it was clear from the get-go that the solution is negative because otherwise the cube would be greater than the square and hence the difference would be negative!

X+3=2 ; X=2-3; X=-1 (Bcz power are same) Very simple

Logb10 24/logb10 6 = logb10 4 or 1.7 he just showed us how to find it without an calculator.

cringe ass video, you are just taaking log and calling it nIcE oLyMpIaD QuEsTiOn. Do some real math baby boy. This video infuriates me to no ends.

I would have thought a Harvard candidate would be able to simplify (1/3)^(1/3) to 1/∛3 in a single step? You just take the cube root of the numerator and denominator. To my mind, that is the simplest way to write the expression. What is gained by writing it as ∛(3^2)/3? That seems more complicated than the original form. But in any case you can get from 1/∛3 to ∛(3^2)/3 in a couple of extra steps.

6^x=24 X=log 24 base 6 Or X=1+log 4 base 6

X= log4

Very simple. a = -5 (minus five)

Here's a method to remember when, like here, you need to find a & b, given their sum (S = a+b) & product (P = ab). If you form the product of these binomials to make the quadratic equation, (x-a)(x-b) = 0, you get x² - (a+b)x + ab = x² - Sx + P = 0 Solve that by any quadratic solution method to get a & b. In the present case, that will be x² - 12x + 66 = 0 The discriminant for this is < 0, so the solutions will be two conjugate complex numbers. Let's use "complete the square." To do that here, that "66" needs to become (½·12)² = 36, so just subtract 30 from both sides: x² - 12x + 36 = -30 x - 6 = ±√(-30) x = 6 ± i√30 Finally, check that the sum & product are the given values. As usual, this is left to the student. Fred

It helps if you explain why you are multiplying by the (√3 - √12) rather than just do it. The reason why we choose to do it is explicitly to use the (a+b)(a-b) = a^2-b^2 equality in order to get rid of the square roots in the divisor. Yes, you show that what's happens, but it is much, much better to explain why you are doing it in the first place. So it would be much better to explain the thinking first, and not make it just appear as the result as if by magic.

Можно короче: а^2 ( 1 - a ) = 150 = (25)(6), то есть 1 - а = 6 и а = - 5 . А комплексные корни можно найти потом из квадратного уравнения.

√x=5+z and √y=5-z; (5+z)(5-z)=10; z^2=15; z=±√15; x=(5±√15)^2 and y=(5∓√15)^2

マイナス5

∛9/3

8 minutes for an operation I did in 5 seconds in my head. lmao

If you can guess to split 150 into 25 and 125, than you should get the answer a= -5 without any calculation 😂

I did it this way: (v3-v12)/(v3+v12)= (v3-2v3)/(v3+2v3)= -v3/(3v3)= -1/3

6 🎉4🎉

1) for a > 1, the equations does not hold 2) for 0 <= a <= 1, the equation does not hold 3) for -1 <= a < 0, the equation does not hold 4) find a=-5 is a root the other part is easy

b = 81; k = 30 b^sin²(x) + b^cos²(x) = k b^(sin²(x)+cos²(x)) + (b^cos²(x))² = k*b^cos²(x) (b^cos²(x))² - k*b^cos²(x) + b^1 = 0 b^cos²(x) = (k/2) ± √((k/2)² - b) 81^cos²(x) = 15 ± √(225 - 81) 3^(4*cos²(x)) = 15 ± 12 = 3, 27 4*cos²(x) = 1, 3 cos(x) = ±1/2, ±√3/2 x = π/6 + k*π/2, π/3 + k*π/2, k ∈ ℤ = (3*k + 1)*π/6, (3*k + 2)*π/6, k ∈ ℤ x = ±π/6 + k*π, ±π/3 + k*π, k ∈ ℤ Unique solutions: x = ±30⁰, ±60⁰, ±120⁰, ±150⁰

Where's the "trick"? He just explores possible combinations of x and y, and if you're doing that it's much quicker to simply subtract 36 from squares of x and seeing which give a square as y squared (so x= 6 gives y=0 and x = 10 gives y = 8).

a=-5 not in 10 seconds but in 20 seconds cos I am not mathematician, I am just ordinary person.

Since you guessed one root y=5 then divide the polynomial (yyy -yy-100)/(y-5) = yy+4y+20 and solve the quadratic equation for the other two roots. Will save you 10 minutes out of 11:47.

asnwer=(1/+/4 ) (1-/4) isit

感覚的に5と解る。

Its the simplest explanation i've seen on KZbin! Thank you very much!!

I exposant 2= -1

The second method actually is virtually the same as the first one. The only difference is that you use the special ln instead of the general log. And writing x = 2.4650 is actually wrong. x is not _equal_ to that number, that is only an _approximation_. (Already saying that ln(15) "is equal to" 2.7081 is wrong.)

Great sir

this is a joke right 💀

a^2(a-1)=-5*5*3*2 thus a=-6

Just intuitively: x=6 y=4

Pretty sure this guy added harvard as a SEO tactic because if this high school level index math is harvard interview level, then i am fuckin Albert Einstein.

Ans: -5

It took me TWO seconds to get to 10 and 8.