Question

what is the answer of this question​

Answer 1

Step-by-step explanation:

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Answer 2

alpha , beta and gamma are zeros of x³ - 12x² + 44x + c and the three zeros are in AP.

Let alpha = a - d , beta = a and gamma = a + d

now, sum of roots = alpha + beta + gamma = -(-12)/1 = 12

=> (a - d) + a + (a + d) = 12

=> 3a = 12

=> a = 4

sum of products of two consecutive roots = alpha.beta + beta.gamma +gamma.alpha

= 44

=> (a -d)a + a(a + d) + (a -d)(a + d) = 44

=> a² - ad + a² + ad + a² - d² = 44

=> 3a² - d² = 44

=> 3(4)² - d² = 44

=> - d² = 44 - 48 = -4

=> d = ±2

hence, a - d = 6 , a = 4, a + d = 2

so, alpha = 6, beta = 4 and gamma = 2

[note :- you can assume , gamma= 6,beta= 4 and alpha = 2 ]

now, product of roots = alpha.beta.gamma= c

=> 6 × 4 × 2 = c

=> c =48

hence, value of c = 48

Hope it helps

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