Q.6) Quadratic Equation

Q : Solve for x ; (x² - 3x + 2)² - 3(x² - 3x + 2) + 2 = x

SOLUTION :
Answer ► x = { 0, 2, 2 ± √2 }


Given : (x² - 3x + 2)² - 3(x² - 3x + 2) + 2 = x

now, (x² - 3x + 2) = (x - 2)(x - 1)
let (x - 2) = y, therefore, (x - 1) = (y + 1)
so, the equation reduces to;
[y(y + 1)]² - 3y(y + 1) - y = 0

=> [y(y + 1)]² - 3y² - 3y - y = 0
=> y² (y² + 2y + 1) - 3y² - 4y = 0
=> y [y^3 + 2y² + y - 3y - 4] = 0
=> y [ y^3 + 2y^2 - 2y - 4] = 0
=> y (y² - 2)(y + 2) = 0
=> y = 0; y = -2; y = ±√2

therefore, x - 2 = 0 ; x - 2 = -2; x - 2 = ±√2
=> x = 2 , x = 0, x = 2 ± √2 ► (Answer)

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