How do you solve #2u ^ { 2} = u + 6#?

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Answer 1 · 2018-04-24T12:25:43

#4# and #-3#

#=> 2u^2 = u + 6#

Rearrange

#=> 2u^2 - u - 6 = 0#

It’s in the form of #au^2 + bu + c = 0#

Where

Use quadratic fromula to find #u#

#u = (-b +- sqrt(b^2 - 4ac))/(2a)#

#u = (-(-1) +- sqrt((-1)^2 - (4 × 2 × (-6))))/(2 × 2)#

#u = (1 +- sqrt(1 + 48))/4#

#u = (1 +- sqrt(49))/4#

#u = (1 +- 7)/4#

Two possible values of #u# are

#u_1 = (1 + 7)/2 = 8/2 = 4#

#u_2 = (1-7)/2 = (-6)/2 = -3#