Question

if the zeros of the polynomial x³-3x²+x+1 are a/r,a and ar,then find the value of a

Answer 1

hope this helps you.....

Answer 2

Answer:

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!!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!!