Question

find the zeroes of Pqx^2+(pr+qs)x+rs and verify the relationship between the zeroes and the coefficient?​

Answer 1

Solution:

i )pqx²+(pr+qs)x+rs

=pqx²+prx + qsx + rs

= px(qx+r)+s(qx+r)

= (qx+r)(px+s)

To find zeroes , we must take

qx+r=0 or px+s = 0

=> x = -r/q or x = -s/p

Therefore,

-r/s , -s/p are two zeroes of

given polynomial.

ii ) Compare pqx²+(pr+qs)x+rs

with ax²+bx+c , we get

a = pq , b =( pr+qs) , c = rs

iii ) sum of the zeroes

= -r/q - s/p

=- (pr+sq)/pq

= -(x-cofficient )/(x²-coeffcient)

iv ) product of the zeroes

= (-r/q)(-s/p)

= rs/pq

= constant/(x²-coefficient)

••••

Answer 2

This is the step by step explanation.

hope this will help you.

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