Question

if a minus b a and a + b are the zeros of the polynomial f bracket x is equal to x cube minus 3 X square + X + 1 then the value of A and B respectively are​

Answer 1

$\huge{\boxed{\underline{\bold{Polynomial}}}}$

$\huge{\boxed{\underline{\bold{Rules}}}}$

• a + (a + b) + (a - b) = - Coefficient of x²/Coefficient of x³
• a(a+b)+(a+b)(a-b)+a(a-b) = Coefficient of x/Coefficient of x³
• a(a + b)(a - b) = - Constant Term/Coefficient of x³

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⇒f(x) = x³ - 3x² + x + 1

Zeros are a, a - b and a + b i.e., in AP.

⇒a + (a + b) + (a - b) = - (- 3)/1

⇒3a = 3

$\Longrightarrow{\large{\boxed{\bold{a = 1}}}}$

⇒a(a+b)+(a+b)(a-b)+a(a-b) = 1

⇒a² + ab + a² - b² + a² - ab = 1

⇒3a² - b² = 1

⇒3(1)² - 1 = b²

$\Longrightarrow{\large{\boxed{\bold{ \pm \sqrt{2} = b}}}}$

$\mathsf{\underline{Hence\:a\:is\:1\:and\:b\:is\: \pm \sqrt{2}}}$

Answer 2

Answer:

Given that;

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

Solution:

We know that-

Sum of zeroes = -b/a

⇒ (a - b) + a + (a + b) = -(-3)/1

⇒ 3a = 3

⇒ a = 3/3

⇒ a = 1

Product of zeroes = c/a

⇒ (a - b) * (a + b) * a = 1

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

⇒ 1² - b² * 1 = 1

⇒ 1 - b² = 1

⇒ b² = 1 + 1

⇒ b² = 2

⇒ b = ± √2

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★ Rules of polynomials-

• sum of zeroes = -b/a
• Product of zeroes = c/a
• Sum of products = -d/a