Japanese | A Nice Algebra Problem | Math Olympiad

[@philglover5110]

I tried this one and was sure I had made a silly mistake when I got the complex solutions - they normally seem to be a lot simpler numbers than that!

[@lrwmasa]

(x²+5x+5-1)(x²+5x+5+1)=120 //let x²+5x+5=a (a-1)(a+1)-120=a²-11=(a+11)(a-11)=0 (x²+5x+16)(x²+5x-6)=(x²+5x+16)(x+6)(x-1)=0

[@brianwade4179]

Multiply it out completely, collect terms, and then move the 120 to the LHS. This gives us x^4 + 10x^3 + 35x^2 + 50x - 96. From inspection of the original problem, or from noticing that the sum of the coefficients in this polynomial is zero, we know x=1 is a root, so (x-1) is a factor. By polynomial division, the remaining polynomial is x^3 + 11x^2 + 46x + 96. From Rational Root Theorem, root candidates are +- (1, 2, 3, 4, 6, 8, ..., 96). Because of those big exponents, try smaller candidates first. x=-6 works, so (x+6) is a factor. By polynomial division, the remaining polynomial is x^2 + 5x + 16. Complete the square: x^2 + 5x + 25/4 = -16 + 25/4 = -39/4 aka (x+5/2)^2 = -39/4 aka x+5/2 = +-sqrt(-39)/2 aka x = -5/2 +- (1/2)sqrt(39)i.

[@robertveith6383]

On the second line you are missing "= 0, because you are dealing with an *equation,* not just a polynomial.

[@NarinderPal-s3q]

1

[@에스피-z2g]

Solution by insight 2×3×4×5=120 x=1

[@robertveith6383]

That is just one of the solutions.