Question

find the zeroes of the polynomial x^3 -21x^2+138x -120 if it is given that zeroes are in arithmetic progression

Answer 1

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

f(x) = x³ - ax² + bx - c

Let the zeroes be

n - d, n, n + d

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

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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Answer 2

$\huge{ \green{ \mathfrak{ \underline{Answer}}}}$

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$\boxed{\red{\bold{Let \: the \: zeroes\: are:}}}$

$(a-d) , a , (a+d) \; \; with\: common\\ difference\: +d$

$\boxed{\red{\bold{Relations \:of\:sum\: of\:zeroes:}}}$

$\implies (a-d) + a + (a+d) = \displaystyle{\frac{-b}{a}}$

$\implies a-d+a+d=\displaystyle{\frac{-(-21)}{1}}$

$\implies 3a = 21$

$\implies a = 7$

$\boxed{\red{\bold{Using\:relations\:of \:sum\: of \:product \:of \:zeroes:}}}$

$\implies (a-d)a + a(a+d) + (a-d)(a+d)= \displaystyle{\frac{c}{a}}$

$\boxed{\red{\bold{Putting \: a=7:}}}$

$\implies (7-d)7 + 7(7+d) + (7-d)(7+d)= \displaystyle{\frac{138}{1}}$

$\implies 49 - 7d + 49 + 7d + 49 - d^2 = 138$

$\implies 147 - d^2 = 138$

$\implies d^2 = 138 - 147$

$\implies d = \pm \sqrt{9}$

$\implies d = \pm 3$

$\boxed{\red{\bold{Putting \: a= 7 \:and \:d=3:}}}$

$Zeroes\: are : (7-3) , 7 , (7+3)$

$\implies 4 , 7 , 10$

$\boxed{\red{\bold{Putting\: a=7 \: and \: d= -3:}}}$

$Zeroes \: are : [7-(-3)] , 7 , [7+(-3)]$

$\implies (7+3), 7 , (7-3)$

$\implies 10 , 7 , 4$

$\boxed{\red{\bold{If \: zeroes\: are\: \alpha , \beta ,\gamma:}}}$

$\alpha = 4$

$\beta = 7$

$\gamma = 10$

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