Question

If the zeros of the polynomial x3 - ax2 + bx - c are A.P, then show that 2a3 - 9ab + 27c= 0.



Also, what is A.P?

Answer 1

Answer:

Step-by-step explanation:

f(x) = x³ - ax² + bx - c

Let say zeroes are  

n-d , n , n + d

Sum of zeroes = 3n

Sum of Zeroes = -(-a)/1 = a

3n = a

n = a/3

n is the zero

f(n) = 0

n³ - an² + bn - c = 0

(a/3)³ - a(a/3)² +ba/3 - c = 0

multiplying by 27 both sides

=> a³ - 3a³ + 9ab - 27c = 0

=> -2a³ + 9ab - 27c = 0

Changing sign

=> 2a³ - 9ab + 27c = 0

QED

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

=> (a/3)(a/3 + a/3)  + (a/3)² - d² = b

=> 3(a/3)² - d² = b

=> d² = a²/3 - b

=> d = ±√(a²/3 - b)

AP is

a/3 +√(a²/3 - b) , a/3  , a/3 -√(a²/3 - b)

a/3 -√(a²/3 - b) , a/3  , a/3 + √(a²/3 - b)

Answer 2

$\huge{\underline{\boxed{\bf{\blue{Answer:-}}}}}$

f(x) = x³ - ax² + bx - c

Let the zeroes be

n - d, n, n + d

$\huge{\underline{\boxed{\bf{\blue{Explainationtion:-}}}}}$

Sum of zeroes = 3n

Sum of Zeroes = - (-a)/1 = a

3n = a

n = a/3

n is the zero

f(n) = 0

n³ - an² + bn - c = 0

(a/3)³ - a(a/3)² +ba/3 - c = 0

Multiplying by 27 both sides

=> a³ - 3a³ + 9ab - 27c = 0

=> -2a³ + 9ab - 27c = 0

Changing sign

=> 2a³ - 9ab + 27c = 0

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

=> (a/3)(a/3 + a/3)  + (a/3)² - d² = b

=> 3(a/3)² - d² = b

=> d² = a²/3 - b

=> d = ±√(a²/3 - b)

Therefore, AP is

a/3 +√(a²/3 - b) , a/3  , a/3 -√(a²/3 - b)

a/3 -√(a²/3 - b) , a/3  , a/3 + √(a²/3 - b)

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