Question

Plz answer above question...

Don't post irrelevant answers...
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Answer 1

$\huge \mathfrak \pink{ÀÑẞWËR}$

Step-by-step explanation:

α,β and γ are zeros of cubic polynomial and are in AP.

So, Let β=a ; α=a−d & γ=a+d

$polynomial = {x}^{3} - 12 {x}^{2} + 44x + c$

$sum \: of \: roots = \frac{ - ( - 12)}{1} = 12$

$so. \: a - d + a + a + d = 12$

$3a = 12$

$a = 4$

$\blue{sum \: of \: products \: of \: tow \: consecutive \: roots = 44}$

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

${a}^{2} - ad + {a}^{2} + ad + {a}^{2} - {d}^{2} = 44$

$3 {a}^{2} - {d}^{2} = 44$

$3 {(4)}^{2} - {d}^{2} = 44$

${d}^{2} = 48 - 44 = 4$

$d = + - 2$

$so. \alpha = a - d = \frac{4 - 2}{4 + 2} = \frac{2}{6}$

$\beta = 4$

$\beta = a + d = \frac{4 + 2}{4 - 2} = \frac{6}{2}$

$so. \: product \: ( - c) = 2 \times 4 \times 6 \times = - 48$

$\red{i \: hope \: it \: helpfull \: for \: you}$

Answer 2

Answer:

by the help of left hand rule move your thumb in force dorecti middle finger in current direction and you can see your last fingure shows you direction of current