If the roots of #2x^2+4x-1=0# are #a# and #b#, find #a^2+b^2#?

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Answer 1 · 2016-12-11T16:34:20

#a^2+b^2=5#

Sum of roots in a qudratic equation #px^2+qx+r=0# is #-q/p# and product of roots is given by #r/p#.

If roots of #2x^2+4x-1=0# are #a# and #b#, we have

we have #(a+b)=-4/2=-2# and

#ab=-1/2#

Hence #a^2+b^2=a^2+2ab+b^2-2ab#

= #(a+b)^2-2ab#

= #(-2)^2-2×(-1/2)#

= #4+1#

= #5#