Zeros of the polynomial x3-3x2+x+1are a/r,a and ar then the value of a is.....
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Step-by-step explanation:
Clearly, sum of the roots is,
a+(a-b)+(a+b)= -(-3)/1=3
=> 3a= 3
=> a= 1.
Again, product of roots is,
a(a-b)(a+b)= -1/1= -1
=>(1-b)(1+b) = -1 [since, a=1]
=> 1- b^2 = -1
=> b^2 = 2
=> b= √2, -√2
Thus, we get, a= 1 & b= √2, -√2.
Hope, it'll help!!
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