A Nice Algebra Problem | Math Olympiad | A Nice Radical Equation

[@quinty.support]

x = √(3x + √4x) Consider squaring of both sides.... → (x)² = √(3x + √4x)² → x² = 3x + √4x → x² ➖ 3x = √4" → Consider squaring of both sides again..... => => (x² ➖ 3x)² = (√4x)² → [Recall (a ± b)² = a² ± 2ab + b²] => x⁴ ➖ 6x³ + 9x² = 4x => => x⁴ ➖ 6x³ + 9x² ➖ 4x = 0 → [Divide the whole expression by x] → x³ ➖ 6x² + 9x ➖ 4 = 0 [Use the method of factorization by splitting and grouping of terms] → [x³ ➖ 4x² ➖ 2x² + 8x + x ➖ 4] = 0 => x²(x ➖ 4) ➖ 2x (x ➖ 4) + (x ➖ 4) → [Obs: (x ➖ 4) is a common factor] => (x ➖ 4)(x² ➖ 2x + 1) = 0 Check if x = 0 is a solution by multiplying both sides by 0.... → (x ➖ 0)(x ➖ 4)(x² ➖ 2x + 1) = 0 Set each of them equal to 0 by brute... Case I: x = 0..... we wanted to tinker with the expression to check for the solution which indeed is one..... Case II: (x ➖ 4) = 0 → x = 4 Case III: (x² ➖ 2x + 1) = 0 [Recognization/Obs: it is a perfect square.....] => (x ➖ 1)² = x² ➖ 2x + 1 plug it on! → (x ➖ 1)(x ➖ 1) = 0 → which implies... x = 1 when x = 1.... however... x = √(3x + √4x)) => 1 = √(3 + √4)) → 1 ≠ √(3 + 2) = √5 which ≠ { ≠ } therefore... x ≠ 1.... rest all the solutions while plugging in for verification satisfies LHS and RHS..... so the pair of solutions for x are → {0, 4} while 1 is {rejected}

[@SALogics]

Very nice! ❤

[@تلميذكممشوش]

❤❤❤❤

[@SALogics]

Very nice! ❤

[@key_board_x]

x = √[3x + √(4x)] x² = 3x + √(4x) x² - 3x = √(4x) (x² - 3x)² = 4x x⁴ - 6x³ + 9x² = 4x x⁴ - 6x³ + 9x² - 4x = 0 x.(x³ - 6x² + 9x - 4) = 0 First case: x = 0 Second case: (x³ - 6x² + 9x - 4) = 0 x³ - 6x² + 9x - 4 = 0 → the aim, if we are to continue effectively, is to eliminate terms to the power 2 Let: x = z - (b/3a) → where: b is the coefficient for x², in our case: - 6 a is the coefficient for x³, in our case: 1 x³ - 6x² + 9x - 4 = 0 → let: x = z - (- 6/3) → x = z + 2 (z + 2)³ - 6.(z + 2)² + 9.(z + 2) - 4 = 0 (z + 2)².(z + 2) - 6.(z² + 4z + 4) + 9z + 18 - 4 = 0 (z² + 4z + 4).(z + 2) - 6z² - 24z - 24 + 9z + 18 - 4 = 0 z³ + 2z² + 4z² + 8z + 4z + 8 - 6z² - 24z - 24 + 9z + 18 - 4 = 0 z³ - 3z - 2 = 0 ← no term to the power 2 z³ - 3z - 2 = 0 → let: z = u + v (u + v)³ - 3.(u + v) - 2 = 0 (u + v)².(u + v) - 3.(u + v) - 2 = 0 (u² + 2uv + v²).(u + v) - 3.(u + v) - 2 = 0 u³ + u²v + 2u²v + 2uv² + uv² + v³ - 3.(u + v) - 2 = 0 u³ + v³ + 3u²v + 3uv² - 3.(u + v) - 2 = 0 u³ + v³ + (3u²v + 3uv²) - 3.(u + v) - 2 = 0 u³ + v³ + 3uv.(u + v) - 3.(u + v) - 2 = 0 u³ + v³ + (u + v).(3uv - 3) - 2 = 0 → suppose that: (3uv - 3) = 0 ← equation (1) u³ + v³ + (u + v).(0) - 2 = 0 u³ + v³ - 2 = 0 ← equation (2) (1): (3uv - 3) = 0 (1): 3uv = 3 (1): uv = 1 (1): u³v³ = 1 ← this is the product P (2): u³ + v³ - 2 = 0 (2): u³ + v³ = 2 ← this is the sum S u³ & v³ are the solution of the equation: a² - Sa + P = 0 a² - 2a + 1 = 0 (a - 1)² = 0 a - 1 = 0 a = 1 u³ = 1 → u = 1 v³ = 1 → v = 1 Recall: z = u + v Recall: x = z + 2 x = (u + v) + 2 x = 4 Resrat x³ - 6x² + 9x - 4 = 0 → we've just seen that 4 is a root, so we can factorize (x - 4) (x - 4).(x² + βx + 1) = 0 → you expand x³ + βx² + x - 4x² - 4βx - 4 = 0 → you group x³ + x².(β - 4) + x.(1 - 4β) - 4 = 0 → you compare wih: x³ - 6x² + 9x - 4 = 0 For x²: → (β - 4) = - 6 → β = - 2 For x: → (1 - 4β) = 9 → 4β = - 8 → β = - 2 (x - 4).(x² + βx + 1) = 0 → where: β = - 2 (x - 4).(x² - 2x + 1) = 0 (x - 4).(x - 1)² = 0 Summarize: x.(x³ - 6x² + 9x - 4) = 0 x.(x - 4).(x² - 2x + 1) = 0 x.(x - 4).(x - 1)² = 0 Solution = { 0 ; 1 ; 4 }

[@SALogics]

Very nice! I really appreciate that ❤

[@rabotaakk-nw9nm]

2:58 x(x³-6x²+9x-4)=0 => => x(x-1)(x²-5x+4)=0 => => x(x-1)(x-1)(x-4)=0 => => x1=0; x2,3=1; x4=4

[@SALogics]

Very nice! ❤

[@rabotaakk-nw9nm]

6:20 x²-2x+1=0 => => (x-1)²=0; x-1=0; x3,4=1

[@SALogics]

Very nice! ❤

[@cabasantbab]

4

[@SALogics]

Very nice! ❤

[@prollysine]

x=0 , (x-4)(x^2-2x+1)=0 , x=4 , x=(2+/-V(4-4))/2 , x=1 , test , x=0 , 4 , OK , x=1 , not a solu , solu , x=0 , 4 ,

[@assiya3023]

معك حق : 1 ليس حل ولكن أين الخطأ في الإجابة ؟

[@YuriyRusinov]

@@assiya3023 Error is breading theorems of equity, that proof that x must be equal to 0 or greater than 3

[@prollysine]

@@assiya3023 x=1 , not a solu , solu , x=0 , 4 ,

[@SALogics]

Very nice! ❤

[@assiya3023]

@@YuriyRusinov هذا ما شككت فيه أنا أيضا شكرا