A Nice Algebra Problem | Math Olympiad | Find x/y=?

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Find the value of x/y?
How to solve (x+y)/√xy=4

In this video, we'll show you How to Solve Math Olympiad Question A Nice Algebra Problem (x+y)/√xy=4 in a clear , fast and easy way. Whether you are a student learning basics or a professtional looking to improve your skills, this video is for you. By the end of this video, you'll have a solid understanding of how to solve math olympiad exponential equations and be able to apply these skills to a variety of problems.
#matholympiad #maths #math #algebra

[RexxSchneider]

Just set x/y = r, the quantity you're trying to find at the start. It makes everything so much easier. Now we can substitute x = ry and get 4 = (ry + y) / √(ry^2) = y(r+1) / y√r = (r+1) / √r. This can be solved for r, or perhaps it's easier to let s= √r and solve for s.
We then have (s^2 + 1) / s = 4 ⇒ s^2 - 4s + 1 = 0. By quadratic formula, s = (4 ± √(16 - 4) ) / 2 = (4 ± √12) / 2 = 2 ± √3. That gives the answer, r = s^2 = 4 ± 4√3 + 3 = 7 ± 4√3.
To check, we can take y=1 and x = r, which gives ( 7 ± 4√3 + 1 ) / √(7 ± 4√3). But we already know that √(7 ± 4√3) = 2 ± √3, so we get (8 ± 4√3) / (2 ± √3) = 4(2 ± √3) / (2 ± √3) = 4.
Incidentally, at 0:30 you can only insist on x/y > 0 for real numbers, which you didn't state. Also xy cannot be 0, but you falsely claim that xy > 0 implies x>0 and y>0, which is clearly nonsense because x<0 and y<0 also gives xy>0.

[guyhoghton]

{ (√x)² + (√y)² } / (√x)(√y) = 4
If you let _u = √(x/y)_ the equation becomes
_u + 1/u = 4_
⇒ _u² - 4u + 1 = 0_
⇒ _u = 2 ± √3_
∴ *_x/y = u² = 4u - 1 = 7 ± 4√3_*

[mrinaldas]

Let x/y=k
Devide numerator and denominator by y
(1+k)/sqrtk=4
(1+k)^2/k=16
1+2k+k^2=16k
Solve the quadratic eqn to get the values of k which are as derived.

[larry]

The first thing I did was square both sides
This resulted in (x^2+y^2+2xy)/xy=16
Separating the fractions
(X^2)/xy+(y^2)/xy+2xy/xy=16
Canceling out variables:
X/y+y/x+2=16
X/y+y/x=14
Substituting a=x/y
A+1/a=14
Multiplying both sides by a
a^2+1=14a
a^2-14a+1=0
by quadratic formula a=7(+-) 4 sqrt 3
a=x/y so x/y=7(+-)4 sqrt 3

[ArwindSah]

√ху≠0 →х≠0 and y ≠0
Divide numerator and denumerator of right hand side by y we get
(x/y +1)/(√(x/y) =4
Let √(x/y) = t. : x/y =t²
(t²+1)/t=4→
t²-4t +1 =0
t =2±√3
x/y= t² = 4t-1 = 7±4√3

[andreasglaser]

x^2+2xy+y^2=16xy

x^2-14xy+y^2=0

x1, 2=7y+-sqrt(49y^2-y^2)

x1, 2=y(7+-4sqrt(3))

x1, 2/y=7+-4sqrt(3)

[فیروزاهنگری]

(x+y)^2=16xy=>x^2-14yx+y^2=0
x=(14y+-8y√3)/2=7y+-4√3y
x=y(7+-4√3)=>x/y=7+-4√3

[허공]

xy ≥ 0 does not mean x ≥ 0 and y ≥ 0. (possibly x ≤ 0 and y ≤ 0)
since x + y = 4√xy ≥ 0, x ≥ 0 and y ≥ 0.

[のえるん-yy]

t^2 - (x+y)t + xy = 0
t^2 - 4(√(xy))t + xy = 0