Question

If the zeroes of the polynomial x³ – 3x² + x + 1 are a – b, a, a + b, find a and b

Answer 1

Since, (a - b), a, (a + b) are the zeroes of the polynomial x³ – 3x² + x + 1.

Therefore, sum of the zeroes = (a - b) + a + (a + b) = -(-3)/1 = 3

⇒ 3a = 3 ⇒ a =1

∴ Sum of the products of is zeroes taken two at a time = a(a - b) + a(a + b) + (a + b) (a - b) =1/1 = 1

a² - ab + a² + ab + a² - b² = 1

⇒ 3a² - b² =1

Putting the value of a,

⇒ 3(1)² - b² = 1

⇒ 3 - b² = 1

⇒ b² = 2

⇒ b = ±`sqrt2`

Hence, a = 1 and b = ± `sqrt2`

Answer 2

p(x) = x³ - 3x² + x + 1

Zeroes are a - b, a + a + b

Comparing the given polynomial with px³ + qx² + rx + 1, we obtain

p = 1, q = - 3, r = 1, t = 1

Sum of zeroes = a - b + a + a +b

`(-q)/p = 3a`

`(-(-3))/1 =3a`

3 = 3a

a = 1

The zeroes are 1-b, 1, 1+b

Multiplication of zeroes = 1(1-b)(1+b)

`(-t)/p=1-b^2`

`(-1)/1 = 1 - b^2`

1- b² = -1

1 + 1 =  b²

b = ±`sqrt2`

Hence a = 1 and b = `sqrt2` or - `sqrt2`