Question

URGENT:- if (a+b), a, (a-b) are zeroes of polynomial x^3-6x^2+11x-6 then find a and b

Answer 1

$p(x) = {x}^{3} - 6 {x}^{2} + 11x - 6$
The zeroes of p(x) are (a-b), a, (a+b).

Sum of roots:

$a - b + a + a + b = 6 \\ 3a = 6 \\ a = 2$
Thus a=2.

Now,

Product of roots:

$a(a - b)(a + b) = 6 \\ 2(2 - b)(2 + b) = 6 \\ 4 - {b}^{2} = 3 \\ {b}^{2} = 1 \\ b = + 1 \\ b = - 1$
Therefore a=2, b=1 or b= -1.

Answer 2

Answer:

a = 2, b = ±1

Step-by-step explanation:

Given Equation is x³ - 6x² + 11x - 6.

On comparing with ax³ + bx³ + cx + d.

We get a = 1, b = -6, c = 11, d = -6.

Let zeroes are α = a + b, β = a, γ = a - b.

(i) Sum of zeroes:

α + β + γ = -b/a

⇒ a + b + a + a - b = 6

⇒ 3a = 6

⇒ a = 2.

(ii) Product of zeroes:

αβ + βγ + γα = c/a

⇒ (a+b)(a) + (a)(a-b) + (a-b)(a+b) = 11

⇒ a² + ab + a² - ab + a² - b² = 11

⇒ 3a² - b² = 11

⇒ 3(2)² - b² = 11

⇒ 12 - b² = 11

⇒ -b² = 11 - 12

⇒ -b² = -1

⇒ b = 1

Therefore, the values are a = 2 and b = ±1

Hope it helps!