Quadratika -2 - Inspired by Sandeep Bhardwaj!
⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ x 1 2 + p x 1 + q = x 2 x 2 2 + p x 2 + q = x 3 x 3 2 + p x 3 + q = x 1
Let p , q be real numbers with α < β be the roots of the equation x 2 + ( p − 1 ) x + q = 0 .
What is the maximum number of solutions of the system of equations above where x 1 , x 2 , x 3 ∈ [ α , β ] is?
The answer is 2.
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1 solution
Your solution is ignoring the fact that on verification of the equations, you will arrive at the condition that x1=x2=x3. Thus out of 8 only 2 solutions will be valid.
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Why shall they be equal......x1,x2,x3 are the roots of the equation and there are 8 total possibilities as x1 can choose from alpha and beta.....similarly for others
As explained by Yash, all that you have shown is x 1 ∈ { a , b } . This implies that there is a maximum of 2 3 = 8 solutions. We still have to verify if any of these work.
Note that if x 1 = a , then x 2 = x 1 2 + p x 1 + q = a and likewise x 3 = x 2 2 + p x 2 + q = a . In particular, ( x 1 , x 2 , x 3 ) = ( a , b , b ) is not possible. There are indeed only 2 solutions.
Taking f(x) = x^2 +(p-1)x +q.....
Let the roots be a,b....
(Where a= alpha, b=beta......which i couldnt type)
Then f(a)=0,f(b)=0...
Now if we add all three equations of x1,x2,x3.....and rearranging them....we get..
Sum = f(x1) + f(x2) + f(x3) =0
(Where f(x1)= (x1)^2 + (p-1)(x1) + q.....and so on)
Now since x1,x2,x3 belongs to [a,b]....
The above Sum =0 iff x1,x2,x3 are either a or b.......
Hence there are 2 options each for x1,x2,x3......so total 8 possibilities...
(Sum =0 iff x1,x2,x3 are either a or b ....because in the interval...
X belongs to (a,b)...
f(x)<0.....as coefficient of x^2 is positive....
So if x1,x2,x3 belongs to [a,b]-{a,b}....
Then f(x1)<0,
f(x2)<0,f(x3)<0.....
So sum=0 is never possible if x1,x2,x3 belongs to [a,b]-{a,b}....
Hence x1,x2,x3 are either a or b).