Question

If (a - b), a and (a + b) are the zeroes of the cubic polynomial f(x)= x³ – 3x² + x + 1, then a =

Answer 1

Given :-

• (a - b), a and (a + b) are the zeroes of the cubic polynomial f(x)= x³ – 3x² + x + 1 ..

To Find :-

• value of a ?

Concept used :-

• The general form of a cubic Equation is ax³ + bx² + cx + d = 0 where a, b, c and d are constants and a ≠ 0.
• The sum of roots of cubic Equation is given by (-b/a) .

Solution :-

Comparing The given Polynomial f(x)= x³ – 3x² + x + 1 with general form of a cubic Equation is ax³ + bx³ + cx + d = 0 , we get :-

→ a = 1

→ b = - 3

→ c = 1

→ d = 1 .

And, Three Roots are given as (a - b), a and (a + b).

So,

→ sum of Roots = (-b/a)

Putting values we get :-

→ (a - b) + a + (a + b) = -(-3)/1

→ a + a + a - b + b = 3

→ 3a = 3

→ a = 1 (Ans.)

∴ value of a will be 1.

Answer 2

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$\huge\tt{GIVEN:}$

• (a - b), a and (a + b) are the zeroes of the cubic polynomial f(x)= x³ – 3x² + x + 1

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$\huge\tt{TO~FIND:}$

• The value of A

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$\huge\tt{SOLUTION:}$

By comparing the polynomial,

↪a = 1

↪b = -3

↪c = 1

↪d = 1

(a - b), a and (a + b) are the zeroes of the cubic polynomial f(x)= x³ – 3x² + x + 1

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→ sum of Roots = (-b/a)

↪ (a - b) + a + (a + b) = -(-3)/1

↪ a + a + a - b + b = 3

↪ 3a = 3

↪a = 1

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