If roots of the equation #x^2+ax-b=0# are #3# and #-7#, find #a# and #b#?

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Answer 1 · 2017-09-05T09:06:14

#a=4#, #b=21# and equation is #x^2+4x-21=0#

In an equation #px^2+qx+r=0#, sum of roots (solutions) is #-q/p# and product of roots is #r/p#.

Hence in #x^2+ax-b=0#, we have

#-a/1=3+(-7)=-4#

or #a=4#

and #-b/1=3xx(-7)=-21#

or #b=21#

and equation is #x^2+4x-21=0#