please solve it and find the value of X
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Given Equation is x^3 - 6x^2 + 11x - 6 = 0
= > x^3 - x^2 - 5x^2 + 5x + 6x - 6 = 0
= > x^2(x - 1) - 5x(x - 1) + 6(x - 1) = 0
= > (x^2 - 5x + 6)(x - 1) = 0
= > (x^2 - 3x - 2x + 6)(x - 1) = 0
= > (x(x - 3) - 2(x - 3))(x - 1) = 0
= > (x - 2)(x - 3)(x - 1) = 0
= > x = 1,2,3.
Hope this helps!
no1vishesh1pbzj6v:
thank you so much
most welcome
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Equation : x³ - 6x² + 11x - 6 = 0
x³ - 6x² + ( 5 + 6 ) x - 6 = 0
=> x³ - 6x² + 5x + 6x - 6 = 0
=> x³ - ( 1 + 5 )x² + 5x + 6x - 6 = 0
=> x³ - x² - 5x² + 5x + 6x - 6 = 0
=> x²( x - 1 ) - 5x( x - 1 ) + 6( x - 1 ) = 0
=> ( x - 1 ) ( x² - 5x + 6 ) = 0
=> ( x - 1 ) ( x² - 2x - 3x + 6 ) = 0
=> ( x - 1 ) [ x( x - 2 ) - 3( x - 2 )] =0
=> ( x - 1 ) ( x - 2 ) ( x - 3 ) = 0
Hence,
x - 1 = 0
=> x = 1
x - 2 = 0
=> x = 2
x - 3 = 0
=> x = 3
x = 1 or 2 or 3
x³ - 6x² + ( 5 + 6 ) x - 6 = 0
=> x³ - 6x² + 5x + 6x - 6 = 0
=> x³ - ( 1 + 5 )x² + 5x + 6x - 6 = 0
=> x³ - x² - 5x² + 5x + 6x - 6 = 0
=> x²( x - 1 ) - 5x( x - 1 ) + 6( x - 1 ) = 0
=> ( x - 1 ) ( x² - 5x + 6 ) = 0
=> ( x - 1 ) ( x² - 2x - 3x + 6 ) = 0
=> ( x - 1 ) [ x( x - 2 ) - 3( x - 2 )] =0
=> ( x - 1 ) ( x - 2 ) ( x - 3 ) = 0
Hence,
x - 1 = 0
=> x = 1
x - 2 = 0
=> x = 2
x - 3 = 0
=> x = 3
x = 1 or 2 or 3
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