Nice Math Problem | Oxford entrance exam question

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[gabrielepolsinelli]

I simply divided both sides for b. Seems way simpler:
(a/b)+1=(6sqrt(ab))/b
(a/b)+1=6sqrt((ab)/b^2)
(a/b)+1=6sqrt(a/b)
Let (a/b)=x and it's the same second grade equation you have had in the end.

[naimyadally]

Could be more concise. Also, the abc in quadratic formula can be confusing with the initial a and b.

[ragingcamel]

Came for the math.
Loved the handwritting.

[uttiyamajumdar]

a+b=6√ab
⇒√(a/b)+√(b/a)=6 [Dividing both sides by √ab]
⇒x+1/x=6 [Taking √(a/b)=x(say)]
⇒x^2-6x+1=0

∴√(a/b)=3±2√2

Thus, a/b=17±12√2

[barygol]

Don't divide by ab and then do the transformation with m and 1/m and eliminate denominator. Divide by b² and you got it directly being a/b -> x

[faizan_-..]

a+b=6√ab (EQ 1)
Sq both side
(a+b)sq = 36ab
a sq + b sq + 2 ab = 36ab
a sq + b sq = 34ab
a sq + b sq -2ab = 32ab
(a-b) sq = 32ab
Root b/s
a-b =4√2 √ab _(EQ 2)
Add EQ 1 and 2nd
a+b = 6√ab
a-b =4√2 √ab
= 2a=6√ab+4√2 √ab
Subract EQ 1 and 2
2b = 6√ab-4√2√ab
Divide both
2a=6√ab+4√2 √ab÷2b = 6√ab-4√2√ab
a/b = 3+ √2/3-√2
Rationalise
a/b=11+6√2/7

[Kumacv]

You are missing the solution where a=b=0 because dividing by zero is not allowed, and ab can, in fact, be equal to zero.

[Vidarl-_-l]

The background music is really annoying 😢

[Gunner]

Just divide original equation by √ab.
Then it becomes √(a/b) + √(b/a) = 6
Let √(a/b) = t, then √(b/a) becomes 1/t. This will form a quadratic equation in t.
Find the value of t by solving the quadratic and pick the positive value(s).
t² is your answer.

[Sanatanrastra]

Or we can also take square of both sides then it become
(a+b)^2=LHS
(6{ab}^1/2)^2=RHS

So LHS=a^2+2ab+b^2
RHS=36ab
Equating LHs and RHS we get
a^2-34ab+b^2=0
Divide both side by ab we get
a/b-34+b/a
Let x=a/b
Then
Equation gets
x-34+1/x
Multiply again by x
We fet equation
x^2-34x+1=0
Use quadratic formula
{-b+(b^2-4ac)^1/2}/2a
And
{-b-(b^2-4ac)^1/2}/2a
Put value each time
b=34, a=1, c=1
We get
a/b=17+12√2
or
a/b=17-12√2

[العابس]

كان من سهل حاله في دفيقه شيل a وحط مكانه nbهتلقي b هتروح وهيفضل علاقه في مجهول واحد❤

[bobovv]

Recognising this as an arithmetic versus geometric average, (look up pythagorean means), you can solve for sin theta = 1/3

[adolfolezama]

Dude you lost me at M lol
But I think I got back on track by the end lol

[Zbigniew-bu]

Prosta sprawa: podzielić przez sqrt(ab) obustronnie i dostajesz równanie x^2-6x+1=o gdzie x=sqrt(a/b) i dostajesz x=3+ -2*sqrt(2) co daje a/b=17+ - 12*sqrt(2)

[sunilmart]

Instead of dividing it by (ab) you should devide it by b²

[AG-idiv]

Oxford? This is a question from Class 11th NCERT Maths textbook from India

[narendersinghal]

Oxford entrance exam question ❎
class 11 NCERT problem💀

[varduhizadoyan]

a/b+b/b=6√ab/b^2
a/b +1 = 6√a/b a/b=x
(x+1) ^2 = 36x
x^2-34x+1=0

[jackardor]

My question -- this is how I see it:

Problem -- a+b=6sqrt(ab)
Divide all by b -- a/b + b/b = 6sqrt(ab)/b
Reduce -- a/b + 1 = 6sqrt(a/b)
All to LHS - a/b -- 6sqrt(a/b) + 1 = 0
Let m = sqrt(a/b) -- m^2 - 6m + 1 = 0
Quadratic -- m = -(-6)/2 +/- sqrt((-6)^2-4(1)(1))/2
Reduce -- m = -3 +/- 2sqrt(2)
Equivalence -- sqrt(a/b) = -3 +/- 2sqrt(2)
Square both sides => a/b = 17 +/- 12sqrt2

What assumptions (besides that a & b are >0) do I need to make to solve it in this way?

[FernandoLopez-wpbs]

EL PROBLEMA DE SER INTRLIGENTE ES QUE NADA ES PARA SIEMPRE