How do you solve #\frac { x ^ { 2} } { 2} - \sqrt { 2} x - 8= 0#?

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Answer 1 · 2018-03-23T12:16:54

Solution : # x=4sqrt2 , x= -2sqrt2 #

#x^2/2- sqrt2 x-8=0 or x^2/2- sqrt2 x=8 # , multiplying

by #2# on both sides we get ,#x^2- 2sqrt2 x=16 #,

added #2# on both sides to complete square on L.H.S

# x^2- 2sqrt2 x +2 =16+2 # or

#(x -sqrt2)^2 =18 or (x-sqrt2)= +- sqrt18 # or

#(x-sqrt2)= +- 3sqrt2 :. x= sqrt2+- 3sqrt2 # or

# x= sqrt2(1+- 3):. x=4sqrt2 or x= -2sqrt2 #

Solution : # x=4sqrt2 , x= -2sqrt2 # [Ans]