Given that the zeros of the cubic polynomial x cube- 6 X square + 3 X + 10 are the form of a, a + b and a + 2 b for some real number A and B find the value of A and B as well as the zeros of given polynomial and also verify the relationship between zeroes and coefficient
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Here is your answer-:
>> Given That a, a+b, a+2b are roots of
given polynomial,
From this polynomial,
Sum of the roots => a+2b+a+a+b=coefficient of
2 3
x / coefficient of x
=> 3a+3b = -(-6)/1 = 6
=> 3(a+b) = 6
=> a+b = 2 ------- (1) b = 2-a
Product of roots = (a+2b)(a+b)a =
constant/coefficient of x3
= (a+b+b)(a+b)a = -10/1
Placing the value of a+b = 2 in it
=(2+b)(2)a = -10
=(2+b)2a = -10
=(2+2-a)2a = -10
=(4-a)2a = -10
=4a-a2 = -5
=a2-4a-5 = 0
=a2-5a+a-5 = 0
=(a-5)(a+1) = 0
a-5 = 0 or a+1 = 0
a = 5 a = -1
a = 5, -1 in (1) a+b = 2
When a = 5, 5+b = 2 => b =-3
a = -1, -1+b = 2 => b = 3
If a = 5 then b = -3
Or
If a = -1 then b = 3
Thank you.
Here is your answer-:
>> Given That a, a+b, a+2b are roots of
given polynomial,
From this polynomial,
Sum of the roots => a+2b+a+a+b=coefficient of
2 3
x / coefficient of x
=> 3a+3b = -(-6)/1 = 6
=> 3(a+b) = 6
=> a+b = 2 ------- (1) b = 2-a
Product of roots = (a+2b)(a+b)a =
constant/coefficient of x3
= (a+b+b)(a+b)a = -10/1
Placing the value of a+b = 2 in it
=(2+b)(2)a = -10
=(2+b)2a = -10
=(2+2-a)2a = -10
=(4-a)2a = -10
=4a-a2 = -5
=a2-4a-5 = 0
=a2-5a+a-5 = 0
=(a-5)(a+1) = 0
a-5 = 0 or a+1 = 0
a = 5 a = -1
a = 5, -1 in (1) a+b = 2
When a = 5, 5+b = 2 => b =-3
a = -1, -1+b = 2 => b = 3
If a = 5 then b = -3
Or
If a = -1 then b = 3
Thank you.
Anonymous:
(-:Thank You dear
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