How do you solve #9(x + 2)^2 - 7 = 0#?

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Answer 1 · 2016-07-23T13:35:33

#-1.118 " or " x = -2.882#

While it is possible to simply and change this into the form #ax^2 + bx + c = 0#, it can be used as it is.

Compare with #5x^2 = 20, " "rArr x^2 = 4 " "rArrx= +-2#

#9(x + 2)^2 - 7 = 0 " move the constant and divide by 9"#

#(x + 2)^2 = 7/9" square root both sides"#

#x + 2 = +-sqrt(7/9)#

#x = +sqrt7/3-2 " or " x = -sqrt7/3 -2#

=#-1.118 " or " x = -2.882#