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China | Math Olympiad Radical Equation | A nice Square Root Problem | You should know this trick
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- Опубликовано: 13 июл 2024
- China | Math Olympiad Radical Equation | A nice Square Root Problem | You should know this trick
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Sqrt[3x+Sqrt[4x]]=x x=0 x=4
I did it in my head.
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.
Awesome 💯👍😎
I solved it the same way.
@@BruceLee-io9by Bruce Lee, the Kung Fu master, is my idol.
@samueltso1291 Bruce Lee was number one, my friend... even in math, it's worth the "be water my friend"!
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
X={0; 4}
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.
√(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)
@@SidneiMV wrong 👎
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)
@@2012tulioread once more
@@SidneiMV ah now you edited your comment
@@2012tulio exactly