solve 8x^4 +4x^3 -18x^2 +11x -2=0, given it has equal roots.
Answers
To solve :
8x⁴ + 4x³ - 18x² + 11 x - 2 = 0
Solution :
8x⁴ + 4x³ - 18x² + 11 x - 2 = 0
=> 8x⁴ + ( 16 - 12 ) x³ + ( 24 - 6 )x² + ( 12 - 1 )x - 2 = 0
=> 8x⁴ - ( 12 - 16 )x³ - ( 6 - 24 )x² - ( 1 - 12 ) x - 2 = 0
=> 8x⁴ - 12x³ + 16x³ + 6x² - 24x² - x + 12x - 2 = 0
=> 8x⁴ - 12x³ + 6x² - x + 16x³ - 24x² + 12x - 2 = 0
=> x [ 8x³ - 12x² + 6x - 1 ] + 2 [ 8x³ - 12x² + 6x - 1 ] = 0
=> ( x + 2 ) [ 8x³ - 12x² + 6x - 1 ] = 0
=> ( x + 2 ) [ 8x³ - ( 8 + 4 ) x² + ( 2 + 4 ) x - 1 ] = 0
=> ( x + 2 ) ( 8x³ - 8x² - 4x² + 2x + 4x - 1 ) = 0
=> ( x + 2 ) ( 8x³ - 8x² + 2x - 4x² + 4x - 1 ) = 0
=> ( x + 2 )( [ 2x { 4x² - 4x + 1 } - 1 { 4x² - 4x + 1 } ] ) = 0
=> ( x + 2 ) [ ( 2x - 1 ) { 4x² - 4x + 1 } ] = 0
=> ( x + 2 )( 2x - 1 ) [ 4x² - ( 2 + 2 ) x + 1 ] = 0
=> ( x + 2 )( 2x - 1 ) [ 4x² - 2x - 2x + 1 ] = 0
=> ( x + 2 )( 2x - 1 ) [ 2x { 2x - 1 } - 1 { 2x - 1 } ] = 0
=> ( x + 2 )( 2x - 1)( 2x - 1 )( 2x - 1) = 0
=> ( x + 2)( 2x - 1 )³ = 0
=> [ x + 2 ][ (2x - 1 )³ ] = 0
=> [ x + 2 ] = 0 and ( 2x - 1 )³ or (1 - 2x )³ = 0
=> x = -2 and (2x - 1) or ( 1-2x ) = 0
=> x = -2 and x = ½ and x = 2
There are three distinct roots , through this is a bi quadratic polynomial with 4 roots .
Three roots of the polynomial are the same .
The graph of this polynomial is attached above :)
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