if the zeros of polynomial x^3-3x^2+x+1 are a-b,a,a+b.find a and b
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heya!!
refer to given attachment please
Remember some rules :-
for a cubic polynomial ax³ + bx² + cx + d
sum of zeros = -(b/a)
sum of product of zeroes taken two at a time = ( c/a)
product of zeros = (d/a)
refer to given attachment please
Remember some rules :-
for a cubic polynomial ax³ + bx² + cx + d
sum of zeros = -(b/a)
sum of product of zeroes taken two at a time = ( c/a)
product of zeros = (d/a)
Attachments:
abishekmr:
where is the attachment
Answered by
8
Heya,
♦ Polynomial → p( x ) = x³ - 3x² + x + 1
You know =_= there's a hell long process but right now, I'll advise this :
→ p( 1 ) = ( 1 )³ - 3( 1 )² + 1 + 1
= 3 - 3 = 0
=> p( x ) = ( x - 1 )( x² - 2x - 1 )
Now, roots of g( x ) = ( x² - 2x - 1 ) :
→ Roots are : ( 1 + √2 ) , ( 1 - √2 ) [ by Quadratic Formula ]
=> Roots of p( x ) are : ( 1 - √2 ) , 1 , ( 1 + √2 )
=> Comparing, a = 1 ; b = ± √2
♦ Polynomial → p( x ) = x³ - 3x² + x + 1
You know =_= there's a hell long process but right now, I'll advise this :
→ p( 1 ) = ( 1 )³ - 3( 1 )² + 1 + 1
= 3 - 3 = 0
=> p( x ) = ( x - 1 )( x² - 2x - 1 )
Now, roots of g( x ) = ( x² - 2x - 1 ) :
→ Roots are : ( 1 + √2 ) , ( 1 - √2 ) [ by Quadratic Formula ]
=> Roots of p( x ) are : ( 1 - √2 ) , 1 , ( 1 + √2 )
=> Comparing, a = 1 ; b = ± √2
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