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A Nice Math Olympiad Problem: x + y = 20, xy = 44; x, y =? y = 20 - x, xy = x(20 - x) = 20x - x² = 44, x² - 20x = - 44, x² - 20x + 100 = - 44 + 100 (x - 10)² = 56 = (4)(14) = (2√14)², x - 10 = ± 2√14; x = 10 ± 2√14 y = 20 - x = 20 - (10 ± 2√14) = 10 -/+ 2√14 Answer check: x = 10 ± 2√14, y = 10 -/+ 2√14 x + y = (10 ± 2√14) + (10 -/+ 2√14) = 20; Confirmed xy = (10 ± 2√14)(10 -/+ 2√14) = 100 - (4)(14) = 100 - 56 = 44; Confirmed Final answer: x = 10 + 2√14, y = 10 - 2√14 or x = 10 - 2√14, y = 10 + 2√14

Log(-2) = 0.301029996 + 1.36437635 i. So it is not impossible.😎

2 seconds in my head. It was so obvious it did not even need thinking about. x = -2, y= 22. Why all those complicated workings?

Never heard of De Moivre formula?

Thank you

you need to decide whether you need a calculator or not. Your screenshot shows "No Calculator" whereas your problem solution uses it. Good problem though.

Poland Math Olympiad Question: 9^√x = √(243ˣ); x =? 9^√x = √(243ˣ) ≥ 0, x ≥ 0 9^√x = 3^(2√x) = √(243ˣ) = 3^(5x/2), 2√x = 5x/2, 5x - 4√x = (√x)(5√x - 4) = 0 x = 0 or 5√x - 4 = 0, √x = 4/5, x = (4/5)² = 16/25 Answer check: x = 0: 9^√x = 9^0 = 1 = √(243⁰) = √(243ˣ); Confirmed x = 16/25, 2√x = 5x/2: 9^√x = 3^(2√x) = 3^(5x/2) = √(243ˣ); Confirmed Final answer: x = 0 or x = 16/25

I thought you were going to find k! (k factorial) :) . You could divide both sides of 5^x=50 by 25=5^2 to simplify the problem to 5^(x-2)=2. It would be easier to apply log base 5

Wtf, why not just take log5 instead of substitution You end up with a = 1/2 + log5(sqrt(3)) Alternatively a = log5(sqrt(15))

5^2a = 5*3 2a log5 = log5 + log3 Divide by log5 on both sides 2a = 1 + (log 3) base 5 a = 1/2 + log sqrt(3) base 5.

x^0.5=2-2^0.50 -> x=(2+2^0.5)^2=4+2-4*2^0.5=6-4*2^0.5=2(3-2*2^0.5)

let u=k+1 , (u-2)(u^5+2u^4+4u^3+8u^2+16u+32)=0 , u=2 , u^5+2u^4+4u^3+8u^2+16u+32=0 , (u+2)(u^4+4u^2+16)=0 , u= -2 , try ()()=0 , (u^2+2u+4)(u^2-2u+4)=0 , u^2+2u+4=0 , u=(-2+/-V(4-16))/2 , u=-1+i*V3 , -1-i*V3 , u^2-2u+4=0 , u=(2+/-V(4-16))/2 , u= 1+i*V3 , 1-i*V3 , solu , u=k+1, k=u-1 , k= 1 , -3 , -2+i*V3 , -2-i*V3 , i*V3 , -i*V3 ,

3^(2*Vx)=3^(x*5/2) , 2*Vx=x*5/2 , 4Vx=5*x , ()^2 , 16x=25x^2 , /:x , 16=25x , x=16/25 , test , 9^(V(16/25))=~ 5.79955 , V(243)^(16/25)=~ 5.79955 , same , OK ,

x^4=16 , (x-2)(x^3+2x^2+4x+8)=0 , x=2 , x^3+2x^2+4x+8=0 , x^2(x+2)+4(x+2)=0 , (x+2)(x^2+4)=0 , x= -2 , x^2+4=0 , x^2= -4 , x= 2i , -2i , solu , x= 2 , -2 , 2i , -2i ,

There are 2 more (real) solutions, in the intervals (1/2, 1) and (-1/2, 0). So it would be clearer to say "find an x ..." and not "solve".

In my head, 3<x<4, and 3 is much closer than 4. Using a calculator, try m= 3.2 - LHS is too low. m =3.3, LHS ~ 119. RHS ~ 119. Close enough. ... Us engineers are a crude bunch, aren't we?

Unless you require m to be an integer this is a wrong answer. There's another real solution approx. 1.41, and probably multiple complex ones.

Nice video

the leap to introduce the index 1/18 isn't explained. a simple written "e.g." might have convinced me not to switch off. So as far as I am concerned: "case unproven".

Nice video

K is 10 .... 50 + 50 = 100

Russia Math Olympiad Question: 4ˣ = (2x)³²; x =? 4ˣ = (2x)³² > 0, x > 0; [(2x)³²]¹⸍⁽³²ˣ⁾ = (4ˣ)¹⸍⁽³²ˣ⁾, (2x)¹⸍ˣ = 4¹⸍³² 4¹⸍³² = 4⁴⸍¹²⁸ = (4⁴)¹⸍¹²⁸ = 256¹⸍¹²⁸ = [2(128)]¹⸍¹²⁸ = (2x)¹⸍ˣ; x = 128

At 1:28, I would realize the bases are the same and equate the exponents. x^1 = x^(sqrt(x)/2) therefore 1 = sqrt(x)/2. 2 = sqrt(x) x = 4.

x = 1.

x = (√x)^(√x) x = x^(√x/2) log(x) = (√x/2)*log(x) log(x) - (√x/2)*log(x) = 0 log(x)*(1 - √x/2) = 0 log(x) = 0 , 1 = √x/2 x = 1 , x = 4

x=2, 2 seconds to calculate the result.

your verification part is nonsense. you cannot simply change the exponent on the left hand side from -0.5 to 2 just by magic while leaving the right hand side untouched... a = -0.5 is also not the right solution. just plug in a = - 0.5 in your calculator and try to get 4 as a result...

At 08:05, you wrote ⁴√ when it should be √ .

x⁴ + x² - 20 = 0 (x² - 4)*(x² + 5) = 0 x² = 4, -5 x = ±2, ±i*√5

At 03:50, you need to include the ± sign: √(u²) = ±√121 Otherwise, your next line is: u = √121 You cannot insert ± in that line, if the line before it doesn't have the ± sign.

Thumbnail has the sum, not the product: 4*x + 10 = 120 x = 27.5 For the product, we have: (x + 1)*(x + 4)*(x + 2)*(x + 3) = 120 (x² + 5*x + 4)*(x² + 5*x + 6) = 120 (x² + 5*x + 5)² - 1² = 120 (x² + 5*x + 5)² = 121 x² + 5*x + 5 = ±11 (x² + 5*x + 5-11)*(x² + 5*x + 5+11) = 0 x = (-5 ± √49)/2, (-5 ± √-39)/2 x = -6, 1, (-5 ± i*√39)/2

x ≠ 0, (20x²)/(4x) = 5, 5x = 5; x = 1

How this 4x becomes 4^x

You went too far. Solution is much simpler. X Cancels with x2. So it becomes 20x/4 =5 Cross multiple , becomes 20x=20 X =20/20=1

X=sqrtX^sqrtX X²=X^sqrtX sqrtX=2=>X=4

Rubbish

(x-5)^(log(5*(x-5)))=2 , let u=x-5 , u^(log5u)=2 , <<< trick , 2=2^log(5*2) , log(5*2)=log10 , log10=1 , 2=2^1 , >>> , if u^log(5*u)=2^log(5*2) true , when u=2 , u=2 , u=x-5 , 2=x-5 , x=2+5 , solu , x=7 , test , (7-5)^log(5*7-25)=2^log(35-25) , 2^log(10) =2^1 , 2^1=2 , same , OK ,

Why do you write log₅2 as log₅² ? It's confusing, as I'm expecting log₅x² .

2*5ᵏ = 100 5ᵏ = 50 k = log(50)/log(5) k = (1 + log(5))/log(5) k = 1/log(5) + 1 = 1/(1 - log(2)) + 1 ≈ 2.4306765580733... Estimate: k ≈ 1/(1 - .30103000) + 1 ≈ 2.4306765 669485... Accuracy is 8 sig figs given the log(2) estimate.

AnsK=2. K•K•k+K=2•2•2+2=8+2=10.

So the answer is either 2, or something else comletely useless in the normal world.

I got it 4194202 while writing instegram notes not paper

You just wasted 0:25 time by writing the samr thing second time :)

How did you jump to 5^k =2.25?

K = 2; 2^3 + 2 = 10