Thailand Junior Math Olympiad Problem | A Nice Algebra Problem

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[SuperYoonHo]

Thanks!!! i subscribed. Great content and keep it up

[prabhudasmandal]

The easier way is, simplify the given value and the determinant which will be( x+2) /√x={x^2--4) /x and (x^2+4) /x=?.now, let the given value =k, then we get (x+2) =k√x..(A)and (x^2--4) =kx.(B). Now, B/A={x--2) =√x or x={x--2) ^2 or x^2--5x+4=0 or (x--4) (x--1) =0 or x=1, 4.so, putting these values we get finally (4^2+4) /4=5, also (1+4) /1=5. Ans.

[dllxjpy]

Sir, we can do this also with the fundamental theorem of the proportions.If (x+2)/sqrt(x)=(x^2-4)/x=(x^2+x-2)/x+sqrt(x)=(x+2)(x-1)/(x+sqrt(x))=(x+2)(sqrt(x)-1)(sqrt(x)+1)/[sqrt(x)(sqrt(x)+1)]=(x+2)/sqrt(x)*(sqrt(x)-1), which leads us to (x+2)/sqrt(x)=(x+2)/sqrt(x)*(sqrt(x)-1), so (x+2)/sqrt(x)*(s-sqrt(x)) =0, which means (x+2)(2-sqrt(x))=0.In conclusion, we have 2 solutions:x=-2 and x=4.but x>0, either the square root of x doesn't exist.x=4 immediately means that x+4/x=5.

[lgooch]

i just brute forced it and factored a cubic and got 5