Question

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

Answer 1

Answer:

Step-by-step explanation:

a-b+a+a+b=3

3a=3

a=1

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

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

1-b^2=-1

b^2=2

b=✓2 or -✓2

Answer 2

a = 1  

b = ±√2

Step-by-step explanation :

Given :

The polynomial : f(x) = x³ - 3x² + x + 1

Zeroes :  (a - b), a & (a + b).

Where,

• a = 1
• b = -3
• c = 1
• d = 1

Consider the :

• α = ( a - b )
• β = a
• γ = ( a + b )

We know that :

$\bigstar\;\boxed{\sf{\bf{\alpha+ \beta + \gamma = \dfrac{-b}{a}}}}}}\\ \\\bf So,\\ \\\implies \sf ( a - b ) + a + ( a - b ) = \dfrac{-(-3)}{1}\\ \\ \\\implies \sf 3a = 3 \\ \\ \\\implies \sf a = \dfrac{3}{3}\\ \\ \\\therefore \bf a = 1$

Hence, The value of a is 1.

Further, Solving :

$\bigstar\;\boxed{\bf \alpha\beta+\beta \gamma+\gamma \alpha=\dfrac{c}{a}}}}\\\\\bf So,\\ \\\implies \sf a( a - b ) + a( a + b ) + ( a + b )( a - b ) = \dfrac{1}{1}\\ \\ \\\implies \sf a^2 - ab + a^2 + ab + a^2 - b^2 = 1 \\ \\ \\\implies \sf 3a^2 - b^2 = 1\\ \\ \\\implies \sf ( 3 \times 1^2 ) - b^2 = 1\qquad\bf[ \because a = 1 ]\\ \\ \\ \implies \sf 3 - b^2 = 1 \\ \\ \\ \implies \sf b^2 = 3 - 1 \\ \\ \\ \implies \sf b^2 = 2 \\ \\ \\ \because \bf b = \pm\sqrt{2}$