Question

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Answer 1

Answer:

a = 1 ; b = ±√2

Step-by-step explanation:

Given polynomial is f(x) = x³ - 3x² + x + 1 .

Here a = 1 , b = -3 , c = 1 , d = 1 .

Let α = ( a - b ) , β = a and γ = ( a + b ) .

As we know,

→ α + β + γ = -b/a .

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

⇒ 3a = 3 .

⇒ a = 3/3 .

∴ a = 1

And,

→ αβ + βγ + γα = c/a .

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

⇒ a² - ab + a² + ab + a² - b² = 1 .

⇒ 3a² - b² = 1 .

⇒ ( 3 × 1² ) - b² = 1 . { ∵ a = 1 }

⇒ 3 - b² = 1 .

⇒ b² = 3 - 1 .

⇒ b² = 2 .

∴ b = ±√2 .

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Answer 2

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$\green{\large\underline{\underline\mathtt{Given:}}}$ The roots of the polynomial are, (a-b), a & (a+b).

$\red{\large\underline{\underline\mathtt{RTF:}}}$ Value of a

$\green{\large\underline{\underline\mathtt{CONCEPT:-}}}$

Relation between roots (x1,x2,x3) of a cubic equation and its coefficients (p,q,r,s) is given by,

x1 +x2 +x3 = -q/p ---> (1)

x1.x2.x3 = -s/p ------>(2)

$\sf{\underline{\boxed{\blue{\large{\bold{ SOLUTION}}}}}}$

We have, x1 =a-b ; x2 = a ; x3 = a +b

And, p=1 ; q=-3 ; r =1 ; s= 1.

Hence, from equation 1 we have,

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

3a = 3

=>${\boxed{a = 1 }}$

$\purple{\large\underline{\underline\mathtt{Extra info:}}}$

To find value of b:-

From equation 2,

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

$(a {}^{2} - b {}^{2} )a = - 1$

Put value of a=1 ,

$(1 - b {}^{2} ) = - 1$

$b {}^{2} = 1 + 1 = 2$

Hence,

${\boxed{b = \sqrt{2} }}$

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