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China | Math Olympiad Radical Equation | A nice Square Root Problem | You should know this trick

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Super Academy

Super Academy

Ай бұрын

China | Math Olympiad Radical Equation | A nice Square Root Problem | You should know this trick
#algebra #exponent#olympiad #simplification#exam
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Пікірлер: 16
@RyanLewis-Johnson-wq6xs
@RyanLewis-Johnson-wq6xs Ай бұрын
Sqrt[3x+Sqrt[4x]]=x x=0 x=4
@RyanLewis-Johnson-wq6xs
@RyanLewis-Johnson-wq6xs Ай бұрын
I did it in my head.
@samueltso1291
@samueltso1291 Ай бұрын
I have a simpler way. For x^3 - 6x^2 + 9x - 4 = 0, one may easily notice that x -1 is a factor. By applying long division, x^3 - 6x^2 + 9x - 4 = (x - 1) (x^2 - 5x + 4) = 0 For (x^2 - 5x + 4), one may easily notice that it can be further factorized to (x - 1)(x -4) Therefore x^3 - 6x^2 + 9x - 4 = (x - 1) (x^2 - 5x + 4) = (x - 1)^2 (x - 4) = 0. x = 1 or x = 4. Together with x = 0, there are three solutions.
@superacademy247
@superacademy247 Ай бұрын
Awesome 💯👍😎
@BruceLee-io9by
@BruceLee-io9by Ай бұрын
I solved it the same way.
@samueltso1291
@samueltso1291 Ай бұрын
@@BruceLee-io9by Bruce Lee, the Kung Fu master, is my idol.
@BruceLee-io9by
@BruceLee-io9by Ай бұрын
@samueltso1291 Bruce Lee was number one, my friend... even in math, it's worth the "be water my friend"!
@walterwen2975
@walterwen2975 Ай бұрын
Math Olympiad Radical Equation: √(3x + √4x) = x; x = ? x ≥ 0; 3x + √4x = x², √4x = x² - 3x = x(x - 3), 4x = [x(x - 3)]² x²(x² - 6x + 9) - 4x = 0, x(x³ - 6x² + 9x - 4) = 0 x = 0 or x³ - 6x² + 9x - 4 = 0, (x³ - 6x² + 8x) + x - 4 = 0 (x - 4)(x² - 2x) + (x - 4) = (x - 4)(x² - 2x + 1) = [(x - 1)²](x - 4) = 0 x - 1 = 0; x = 1, Double root or x - 4 = 0; x = 4 Answer check: x = 0: √(3x + √4x) = 0 = x; Confirmed x = 1: √(3 + √4) = √5 > 1 = x; Failed x = 4; √(12 + √16) = √(12 + 4) = 4 = x; Confirmed Final answer: x = 0 or x = 4
@2012tulio
@2012tulio Ай бұрын
X={0; 4}
@hiddenintheshadows1903
@hiddenintheshadows1903 27 күн бұрын
sqrt(3x+sqrt(4x)) = x x^2 - 3x = sqrt(4x) x^4 - 6x^3 + 9x^2 = 4x x^4 - 6x^3 + 9x^2 - 4x = 0 (x-4)( x^4 - 6x^3 + 9x^2 - 4x). x^3 - 2x^2 + x x^4 - 4x^3 - 2x^3 + 9x^2 - 4x - 2x^3 + 8x^2 x^2 - 4x x^2 - 4x 0 (x-4)(x^3 - 2x^2 + x) =0 x-4 = 0 x1 = 4 x^3 - 2x^2 + x = 0 x(x^2 - 2x + 1) = 0 x2 = 0 x^2 - 2x + 1 = 0 x^2 - x - x + 1 =0 x(x - 1) - (x - 1) = 0 (x-1)(x-1) = 0 x-1=0 x3 = x4 = 1 Only x1 and x2 are viable solutions for the problem.
@SidneiMV
@SidneiMV Ай бұрын
√(3x + 2√x) = x √x = u => x = u² √(3u² + 2u) = u² 3u² + 2u = u⁴ u⁴ - 3u² - 2u = 0 u(u³ - 3u - 2) = 0 u = 0 => *x = 0* u³ - 3u - 2 = 0 u³ - 8 - 3u + 6 = 0 (u - 2)(u² + 2u + 4) - 3(u - 2) = 0 (u - 2)(u² + 2u + 1) = 0 (u - 2)(u + 1)² = 0 u - 2 = 0 => *x = 4* u + 1 = 0 => u = -1 (not valid)
@2012tulio
@2012tulio Ай бұрын
@@SidneiMV wrong 👎
@SidneiMV
@SidneiMV Ай бұрын
another way √(3x + 2√x) = x 3x + 2√x = x² 2√x = x² - 3x 4x = x⁴ + 9x² - 6x³ x⁴ - 6x³ + 9x² - 4x = 0 x(x³ - 6x² + 9x - 4) = 0 *x = 0* x³ - 6x² + 9x - 4 = 0 x³ - 1 - 3(2x² - 3x + 1) = 0 x³ - 1 - 3(2x - 1)(x - 1) = 0 (x - 1)(x² + x + 1) - 3(2x - 1)(x - 1) = 0 (x - 1)(x² + x + 1 - 6x + 3) = 0 (x - 1)(x² - 5x + 4) = 0 (x - 1)(x - 1)(x - 4) = 0 (x - 4)(x - 1)² = 0 x - 4 = 0 => *x = 4* x - 1 = 0 => x = 1 (not valid)
@SidneiMV
@SidneiMV Ай бұрын
​@@2012tulioread once more
@2012tulio
@2012tulio Ай бұрын
​@@SidneiMV ah now you edited your comment
@SidneiMV
@SidneiMV Ай бұрын
@@2012tulio exactly
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