Question

pls ans to be brainlist

Answer 1

Given f(x) = x^3 - 6x^2 + x + 2.

On comparing with ax^3 + bx^2 + cx + d, 

a = 1, b = -6, c = 1, d = 2

Given that p - q, p and p + q are the zeroes of f(x).

Now,

= > We know that sum of zeroes = -b/a

p - q + p + p + q = -(-6)/1

3p = 6

p = 2.


We know that Product of roots = c/a

= > (p - q)p + p(p + q) + (p + q)(p - q) = 1/1

= > p^2 - pq + p^2 + pq + p^2 - q^2 = 1

= > 3p^2 - q^2 = 1

= > 3(2)^2 - q^2 = 1

= > 12 - q^2 = 1

= > q^2 = 11

$= \ \textgreater \ q = + \sqrt{11} , - \sqrt{11}$



Therefore the value of p = 2, q = +root 11, -root 11.


Hope this helps!

Answer 2

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