What is the solution set for #x^2 - 4x = 8#?

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Answer 1 · 2015-08-15T13:57:54

Refer explanation

#x^2# - 4x - 8 = 0

Examine
#b^2# - 4ac
#-4^2# - (# 4 xx 1 xx -8#) = 16 + 32 = 48 ( positive and not a perfect square . So use formula)

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

x = #(-(-4) +- sqrt((-4)^2 - ( 4 xx 1 xx -8 )))/ (2 xx 1)#

x = #(4+- sqrt((16 - ( -32 ))))/ (2)#

x = #(4+- sqrt(16 + 32 ))/ (2)#

x = #(4+- sqrt(48 ))/ (2)#

x = #(4+- 6.9 )/ (2)#
x = #(4+ 6.9 )/ (2)# = 5.45

x = #(4- 6.9 )/ (2)# = - 1.45