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If x^{4}-8 x^{3}+a x^{2}+b x+16=0 has all positive real roots, then the value of 'a' is

Option: 1

20


Option: 2

8


Option: 3

24


Option: 4

28


Answers (1)

best_answer

Lets roots be p, q, r, s

Sum of roots= p+q+r+s=8----(i)
Product of roots = p.q.r.s=16
Observe that AM of roots =\frac{p+q+r+s}{4}=2

and GM of the roots=\left ( p.q.r.s \right )^{1/4}=2

So AM = GM, which is possible only when p=q=r=s

Using  (i) \Rightarrow 4p=8\Rightarrow p=2\\ \Rightarrow p=q=r=s=2

Now

\\a=p.q+p.r+p.s+q.r+q.s+r.s\\=4+4+4+4+4+4\\=24

Posted by

Ajit Kumar Dubey

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