Nice problem; and a great initial insight. However I got both answers in about the same work just turning the crank.
@StaR-uw3dc
7 ай бұрын
Nice solution. At the point 2(a+b)^2=ab we can write the equation as (a+b)^2-ab/2 = a^2+2ab+b^2-ab/2 = a^2+(3/2)ab+b^2 = (a+(3/4)b)^2 +(7/16)b^2 = 0 and LHS is always non-negative as a sum of two squares. It is equal to 0 only for a=b=0 which is not satisfied by any real x (x cannot be simultaneously equal 9 and 1/9).
@mariosanagnostou3298
7 ай бұрын
another way is if you let a=(3-x^1/2)/(1+x^1/2) and b=(1-3x^1/2)/(1+x^1/2) then observe that a-b=2 next you find a^2+b^2=4+2ab and you know that a^4+b^4=16. After you do (a^2+b^2)^2 and you create an equation with only ab and you find ab=0 or ab=-8 the rest are simple.
@zaman-qn6tm
7 ай бұрын
Sir please Pascal triangle expansion ka use wala ek video banaye
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