300 Followers problem (Not the old one)

Algebra Level 2

x 2 = 4 3 4 3 10 3 x , x = ? \large x-2 = \sqrt{4-3\sqrt{4-3\sqrt{10-3x}}} \quad,\quad x= \, ?


The answer is 3.

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4 solutions

First Last
Dec 19, 2015

It can be seen that x 2 x-2 must be positive, as n \sqrt{n} returns a positive value or zero. Therefore x > = 2 x>=2 . And 10 3 x > = 0 10-3x>=0 to avoid imaginary numbers and must be a perfect square. From this, x x must be 1, 2, or 3. Of these, only 3 \boxed{3} satisfies all conditions.

Why should it be a ​perfect square?

ameer hamza - 5 years, 5 months ago

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Good point. I just assumed it was, but you're right it wouldn't have to be.

First Last - 5 years, 5 months ago

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By solving the quadratic that results your realize it is a perfect square.

Mardokay Mosazghi - 5 years, 5 months ago
Anish Harsha
Dec 19, 2015

x 2 = 4 3 4 3 10 3 x \large x-2 = \sqrt{4-3 \sqrt{4-3 \sqrt{10-3x}}}

Let f ( x ) f(x) = 4 3 x \sqrt{4 -3x} . Equation is f ( f ( f ( x 2 ) ) ) = x 2 f(f(f(x-2))) = x-2 .

Since f ( x ) f(x) is continuous decreasing over its domain, so is f ( f ( f ( x ) ) ) f(f(f(x))) .

So, f ( f ( f ( x ) ) ) = x f(f(f(x)))=x at most one real root (continuous decreasing in LHS and increasing in RHS).

And, since 1 is trivially a solution of f ( x ) = x f(x)=x , it is a solution if f ( f ( f ( x ) ) ) = x f(f(f(x)))=x

Thus, the unique solution is x = 3 \ x=3

Pedro Vitor
Dec 22, 2015

Just do, √4-3 = √1 = 1. Then: x - 2 = √10 - 3x, (x-2)² = (√10-3x)², x² - x - 6 = 0, x' = 3, x" = -2.

Karan Gujar
Dec 20, 2015

Someone plz solve this:In triangle ABC,Angles are in g.p.(A,B,C) with common ratio 2, then find 1/b+1/c-1/a.

lol post this as a note sir.

Mardokay Mosazghi - 5 years, 5 months ago

Hi karan as the values of a, b and c are in gp with a cratio of 2 we can write b and c as 2a and 4a. The sum of angles in a triangle is 180 so the value of a is 25.71, b is 51.43 and c is 102.86. From these three values u can find the value of the required expression

gautham nukala - 5 years, 5 months ago

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