How do you solve #2x ^ { 2} + 6= - 34#?

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Answer 1 · 2016-12-26T13:51:13

#x = isqrt(20)# if we let #i = sqrt(-1)#

First, subtract #color(red)(6)# from each side of the equation to isolate the #x# term and keep the equation balanced:

#2x^2 + 6 - color(red)(6) = -34 - color(red)(6)#

#2x^2 + 0 = -40#

#2x^2 = -40#

Now divide by #color(blue)(2)# to solve for #x^2# and keep the equation balanced:

#(2x^2)/color(blue)(2) = (-40)/color(blue)(2)#

#(color(blue)(cancel(color(black)(2)))x^2)/cancel(color(blue)(2)) = -20#

#x^2 = -20#

Because a number squared is always positive the is no REAL number solution to this problem.

However is we let #i = sqrt(-1)# we can solve as:

#sqrt(x^2) = sqrt(-20)#

#x = sqrt(-1 * 20)#

#x = isqrt(20)#