If a - b , a , a + b are zeros of polynomial x3-3x2 + x + 1 the value of(a+b) is:
Why you didn't gave answer step by step
Im reaallly confused with this answer pliz give it withnice explanation
To find the value of (a + b), we can make use of the fact that the sum of the roots of a polynomial can be determined by the coefficient of the second-to-last term, which is -3 in this case.
Given that the polynomial is x^3 - 3x^2 + x + 1, we can rearrange it to be in the form (x - a)(x - b)(x - c) = 0, where a - b, a, and a + b are the zeros of the polynomial.
Expanding this equation, we get (x^2 - (a + b)x + ab) (x - c) = 0.
Comparing coefficients with the original polynomial, we can equate the corresponding terms:
x^3 - 3x^2 + x + 1 = x^3 - (a + b + c)x^2 + (ab + ac + bc)x - abc.
By comparing coefficients, we find that:
(a + b + c) = 3
ab + ac + bc = 1
abc = -1
Since we are interested in finding the value of (a + b), we can ignore the third equation abc = -1 for now. To solve the first two equations, we can first find the value of c using the equation (a + b + c) = 3, and then substitute it into the equation ab + ac + bc = 1.
Let's solve the equation (a + b + c) = 3:
3 = a + b + c
c = 3 - (a + b)
Now substitute the value of c into the equation ab + ac + bc = 1:
ab + a(3 - (a + b)) + b(3 - (a + b)) = 1
ab + 3a - a^2 + 3b - ab + b^2 = 1
2ab + 3a + 3b - a^2 - b^2 = 1
2ab + 3(a + b) - (a^2 + 2ab + b^2) = 1
2ab + 3(a + b) - (a + b)^2 = 1
Notice that the expression (a + b)^2 appears in the equation, which is equal to (a^2 + 2ab + b^2). We can substitute this back into the equation:
2ab + 3(a + b) - (a^2 + 2ab + b^2) = 1
2ab + 3(a + b) - (a + b)^2 = 1
2ab + 3(a + b) - (a + b)(a + b) = 1
2ab + 3(a + b) - (a + b)^2 = 1
Factoring out (a + b) from both terms, we get:
(a + b)(2 - (a + b)) = 1
Since we are trying to find the value of (a + b), we can solve this equation to obtain the possible values.
Setting (a + b) = t, we have:
t(2 - t) = 1
Expanding this equation, we have:
2t - t^2 = 1
Rearranging the equation, we get:
t^2 - 2t + 1 = 0
This is a quadratic equation, and we can solve it using factoring or the quadratic formula. Factoring this equation, we get:
(t - 1)(t - 1) = 0
As we can see, the equation has one repeated root, t = 1.
Since we set (a + b) = t, we have:
(a + b) = 1
Hence, the value of (a + b) is 1.
(x-(a-b))(x-a)(x-(a+b))
= x^3 - 3ax^2 + (3a^2-b^2)x + ab^2-a^3
If that is identical to x^3-3x^2+x+1 then
-3a = -3
3a^2-b^2 = 1
ab^2-a^3 = 1
a=1 b=±√2