Question

If α,β are the zeros of the polynomial, x2-px +36 and α2 + β2 = 9, then what is the value of p?

Answer 1

Step-by-step explanation:

Solution:-

because a & b are the zeroes of polynomial x2-px+q.......

then (a+b)= -(-p)= p...

(ab)= q

so..... (a2+b2)= (a+b)^2 - 2ab

=p^2 - 2q

Answer 2

Answer:

±9

Step-by-step explanation:

here a=1, b=-p c= 36

$\alpha$+$\beta$= -b/a = -(-p/1)= +p

$\alpha \beta$= c/a = 36/1= 36

$\alpha^{2} +\beta ^{2}$= $(\alpha +\beta)^{2}$ - 2$\alpha \beta$

9=$p^{2}$- 2 x 36

$p^{2}$=81

p=$\sqrt{81}$

p=±9