Math Olympiad | A Nice Algebra Problem | 90% Failed to solve!

[@antonio9633]

Excelente solução 😊

[@vijaymaths5483]

Thank you 😊

[@Quest3669]

Let a and b r 1st n 2nd inner terms then a^2+b^2= 10 &
a-b= -2 gives a= -3;1; b= -1; 3 for a; x^2--x-10= -3x; x gives
X= +-√11+-1 solns.

[@RealQinnMalloryu4]

({x^4 ➖ x^2} ➖ 100/100x^2) ({ 2:27 x^4 +x^2 }➖ 100/100x^2)={(x^2 ➖ 100}/100x^2)+({x^6 ➖ 100}/100x^2)={98/100x^2+94/100x^2}=192/200x^4=10^10^8^12/10^20x^4 ^8^6^6/1^2x^4 2^3^3^3^3/1^2x^4 1^1^1^1^1^1^3/1^2x^2^2 1^3/1x^1^2 3/x^2 (x ➖ 3x+2).

[@KC_Tripathy]

It took time, but i did it my way...

[@walterwen2975]

Math Olympiad: [(x² - x - 10)/10x]² + [(x² + x - 10)/10x]² = 1/10; x =?
x ≠ 0; (x² - x - 10)² + (x² + x - 10)² = (1/10)(10x)², (x² - x - 10)² + (x² + x - 10)² = 10x²
(x² - 10 - x)² + (x² - 10 + x)² = 2(x² - 10)² + 2x² = 10x², (x² - 10)² - (2x)² = 0
(x² - 2x - 10)(x² + 2x - 10) = 0, x² - 2x - 10 = 0 or x² + 2x - 10 = 0
x² - 2x + 1 = (x - 1)² = (√11)², x = 1 ± √11 or (x + 1)² = (√11)², x = - 1 ± √11
Answer check:
[(x² - x - 10)/10x]² + [(x² + x - 10)/10x]² = 1/10
x = 1 ± √11: x² - 2x - 10 = 0, x² - 10 = 2x
[(2x - x)/10x]² + [(2x + x)/10x]² = (1/10)² + (3/10)² = 10/100 = 1/10; Confirmed
x = - 1 ± √11: x² + 2x - 10 = 0, x² - 10 = - 2x
[(- 2x - x)/10x]² + [(- 2x + x)/10x]² = (- 3/10)² + (- 1/10)² = 10/100 = 1/10; Confirmed
Final answer:
x = 1 + √11, x = 1 - √11, x = - 1 + √11 or x = - 1 - √11