Question

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

Answer 1

Answer:

given equation :x^3 -3x^2 +X+1

zeroes are a-b , a and a+b

sun of zeroes :a-b+a+a+b=3

3a=3

a=1

product of zeroes :(a-b)(a)(a+b)=-1

put a = 1

(1-b)(1)(1+b)=-1

1-b^2=-1

1+1=b^2

b=√2

so for the above equation, a=1 and b=√2

Step-by-step explanation:

for cubic equation, sum of roots is -b/a

product of roots is -d/a

Answer 2

Answer:

a = 1 and b = ±√2

Step-by-step explanation:

Given :

The zeroes of the polynomial x³ - 3x² + x + 1 are a - b, a, and a + b

To find :

the values of a and b

Solution :

For a cubic polynomial of the form ax³ + bx² + cx + d , the relation between zeroes and coefficients is given as :

let the zeroes be α, β and γ

• α + β + γ = -b/a
• αβ + βγ + αγ = c/a
• αβγ = -d/a

So,

• a = 1

• b = -3

• c = 1

• d = 1

and

• α = a - b
• β = a
• γ = a + b

Sum of zeroes,

a - b + a + a + b = -(-3)/1

 3a = 3

  a = 3/3

  a = 1

Product of zeroes,

(a - b) (a) (a + b) = -1/1

a( a² - b²) = -1     [ (a + b) (a - b) = a² - b² ]

1(1² - b²) = -1   [ ∵ a = 1 ]

  1 - b² = -1

  b² = 1 + 1

  b² = 2

  b = ±√2

Therefore, a = 1 and b = ±√2