Show that if #p,q,r,s# are real number and #pr=2(q+s)# then at-least one of the equations #x^2+px+q=0# and #x^2+rx+s=0# has real roots?

1 Answer
Jan 8, 2018

Please see below.

Explanation:

The discriminant of #x^2+px+q=0# is #Delta_1=p^2-4q#

and that of #x^2+rx+s=0# is #Delta_2=r^2-4s#

and #Delta_1+Delta_2=p^2-4q+r^2-4s#

= #p^2+r^2-4(q+s)#

= #(p+r)^2-2pr-4(q+s)#

= #(p+r)^2-2[pr-2(q+s)]#

and if #pr=2(q+s)#, we have #Delta_1+Delta_2=(p+r)^2#

As sum of the two discriminants is positive,

at least one of them would be positive

and hence atleast one of the equations #x^2+px+q=0# and #x^2+rx+s=0# has real roots.