Thank you for your effort and work. Even though I didn't fully understand you, the language of mathematics is number one, I understood the equations you wrote on the board. I am Turkish. My English is not at a good level, I hope I wrote it correctly.
Thanks you so much for watching and commenting despite the language barrier sir. You English level is good from your write up here. It is well constructed and I understood every point you pointed out here sir. We love you dearly sir...💖💖💖💖💕💕💕💕❤❤❤😍😍😍😍
My reasoning is that, if neither x or y are greater than 10 they'd both have to equal 10 to make equation 2 correct. But since this would make their combined value greater than 10 so equation 1 would be incorrect. Therefore one of the variables must be greater than 10, but this would mean one of the variables must be less than zero to make equation 1 correct. But if only one of them is a negative number then equation 2 would be incorrect. I therefore concluded that there is no combination of real numbers that solves these equations. And I was right.
when you have quadratic equation z^2-a*z *b=0, you know that a is sum of its roots and b is their product. In our case we know the x and y are roots of z^2-10*z +100=0, from here you get x=5 + 5*i*sqrt(3) and y=5 - 5*i*sqrt(3) or vice versa.
x and y will both have to have real parts 5, so that x+y=10 and the imaginary parts will have to be complements of each other. The modulus of each number will have to be 10 so that xy=100. Therefore the imaginary part is 10^2 - 5^2 = 5sqrt(3) So x and y = 5 +- 5sqrt(3) i
There's no real solution, but nevertheless, you can get the solution as they're the same as this equation: x² -10x + 100 = 0 ∆ = -300 x = 5 + 5i√3 y = 5 -5i√3 Or the opposite
Must all Cambridge entrance math problems be brain racking at all time, sir? There are some questions that are always "give away" amongst tough ones in every admission interview.
C'est un calcul basique utilisé d'innombrables fois en Physique. J'ai souvent observé que les gens qui ne savent pas à quoi sert quelque chose en concluent que ce quelque chose ne sert à rien !
Thank you for your effort and work. Even though I didn't fully understand you, the language of mathematics is number one, I understood the equations you wrote on the board. I am Turkish. My English is not at a good level, I hope I wrote it correctly.
Thanks you so much for watching and commenting despite the language barrier sir. You English level is good from your write up here. It is well constructed and I understood every point you pointed out here sir.
We love you dearly sir...💖💖💖💖💕💕💕💕❤❤❤😍😍😍😍
My reasoning is that, if neither x or y are greater than 10 they'd both have to equal 10 to make equation 2 correct.
But since this would make their combined value greater than 10 so equation 1 would be incorrect.
Therefore one of the variables must be greater than 10, but this would mean one of the variables must be less than zero to make equation 1 correct.
But if only one of them is a negative number then equation 2 would be incorrect.
I therefore concluded that there is no combination of real numbers that solves these equations.
And I was right.
when you have quadratic equation z^2-a*z *b=0, you know that a is sum of its roots and b is their product. In our case we know the x and y are roots of z^2-10*z +100=0, from here you get x=5 + 5*i*sqrt(3) and y=5 - 5*i*sqrt(3) or vice versa.
x and y will both have to have real parts 5, so that x+y=10 and the imaginary parts will have to be complements of each other.
The modulus of each number will have to be 10 so that xy=100.
Therefore the imaginary part is 10^2 - 5^2 = 5sqrt(3)
So x and y = 5 +- 5sqrt(3) i
There's no real solution, but nevertheless, you can get the solution as they're the same as this equation:
x² -10x + 100 = 0
∆ = -300
x = 5 + 5i√3
y = 5 -5i√3
Or the opposite
Resolver:
x+y=10(1)
x.y=100(2)
De (1) despejo y:
y=10-x(3)
Ahora llevo (3) en (2):
x(10-x)=100
Ahora resuelvo para x:
→10x-x²=100
-x²+10-100=0
(-1).(-x²+10-100)=0
→x²-10+100=0
Resuelvo para x:
x=-(-10)±√100-4(1)(100)/2
x=10±√100-400/2
x=10±√-300/2
x=10±√-1.√3.5².2²/2
x=10±10√3i/2
x=5 ±5√3i
Aquí: x'=5+5√3i✓
x''=5-5√3i✓
Calculo de y:
Si x'=5+5√3i
y'=10-(5+5√3i)
y' =10-5-5√3i
y'=5-5√3i✓
y''=10-(5-5√3i)
y"=10-5+5√3i
y"=5+5√3i✓
I wonder why such an easy task would be a part of the application for maths at Cambridge.
Must all Cambridge entrance math problems be brain racking at all time, sir?
There are some questions that are always "give away" amongst tough ones in every admission interview.
brother I learned this in 9th grade here in germany, and you have to do this to enter in cambridge?
Me, too! 😂 However the complex solution usually isn't taught at school. We had it in year 12 or 13 in maths.
(5+5×√−1×√3)+(5−5×√−1×√3)
(5+5×√−1×√3)*(5−5×√−1×√3)
Cambridge 😮
😂😂😂😂😂😂😂
Fake news
that is cute but what is the point, you owe me $i dollars for that 10 minutes
Hahahahahaha.....🤣🤣😂😂🤣🤣😂😂🤣🤣😍😍
Drop your account details then.
C'est un calcul basique utilisé d'innombrables fois en Physique. J'ai souvent observé que les gens qui ne savent pas à quoi sert quelque chose en concluent que ce quelque chose ne sert à rien !
gosh this is awful ...
Thanks a million sir.