Well, from another video method: Vieta theorem We have: sqrt(x)+sqrt(y)=5 (-b) sqrt(xy)=sqrt(x)*sqrt(y)=5 (c) t^2-5t+5=0 t(1,2)=(5+-sqrt(25-20))/2 t1=(5+sqrt(5))/2;t2=(5-sqrt(5))/2 (t1)^2=x1=y2=(5+sqrt(5))^2/4=(15+5sqrt(5))/2 (t2)^2=x2=y1=(5-sqrt(5))^2/4=(15-5sqrt(5))/2 Answer:((15+5sqrt(5))/2;(15-5sqrt(5))/2);((15-5sqrt(5))/2;(15+5sqrt(5))/2)
15 can be written as (30/2) and (30/2)-(15/2) = 15/2 . another small example :- 3 -(3/2) = 3/2. So visually ..it may appear disappeared but it is not actually. And also not just 15 is disappeared along with a minus sign is missing which should be noticed I'm afraid that you asked this question.
Well, from another video method: Vieta theorem
We have:
sqrt(x)+sqrt(y)=5 (-b)
sqrt(xy)=sqrt(x)*sqrt(y)=5 (c)
t^2-5t+5=0
t(1,2)=(5+-sqrt(25-20))/2
t1=(5+sqrt(5))/2;t2=(5-sqrt(5))/2
(t1)^2=x1=y2=(5+sqrt(5))^2/4=(15+5sqrt(5))/2
(t2)^2=x2=y1=(5-sqrt(5))^2/4=(15-5sqrt(5))/2
Answer:((15+5sqrt(5))/2;(15-5sqrt(5))/2);((15-5sqrt(5))/2;(15+5sqrt(5))/2)
Well done!
Thank you so much!
7:30, how 15 disappeared
15 can be written as (30/2) and (30/2)-(15/2) = 15/2
. another small example :- 3 -(3/2) = 3/2.
So visually ..it may appear disappeared but it is not actually.
And also not just 15 is disappeared along with a minus sign is missing which should be noticed
I'm afraid that you asked this question.
you had 15 minus half of it. that's just the other half :)
√x + √y = 5 --- ①
√xy = 5 --- ②
From ①
(√x + √y)² = 5²
x + y + 2√xy = 25 ②
√x(15 - x) = 5
[√x(15 - x)]² = 5²
x(15 - x) = 25
- x² + 15x = 25
x² - 15x + 25 = 0
D = (- 15)² - 4*1*25 = 225 - 100 =125 = 5³ > 0
x = [- (- 15) ± √D ] / 2*1
x = ( 15 ± 5√5 ) / 2
y = 15 - ( 15 ± 5√5 ) / 2
y = (15 -+ 5√5 ) / 2
∴ (x , y) = {(15 + 5√5) / 2 , (15 - 5√5) / 2} , {(15 - 5√5) / 2 , (15 + 5√5) / 2}
Verifying
√xy = √[(15 + 5√5) / 2] * [(15 - 5√5) / 2]
= √[(15 + 5√5)(15 - 5√5) / 4]
= √[(225 - 125) / 4]
= √100/4
= √25
= 5 Pass!