Question

Find the zeroes of the quadratic polynomial pqx+ + (qr+qs)x + rs and verify the relationship between the zeroes
and coefficients.
​

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)