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

Answer 1

Anna Allen

#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#

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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