How do you solve #(x + 6)^{2} = 1331#?

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Answer 1 · 2016-09-21T06:24:57

#x = -6+-11sqrt(11)#

Note first that #1331 = 11^3#

So taking square roots of both sides of the equation and allowing for both positive and negative square roots, we have:

#x+6 = +-sqrt(1331) = +-sqrt(11^3) = +-sqrt(11^2*11)#

#= +-sqrt(11^2)*sqrt(11) = +-11sqrt(11)#

Then subtract #6# from both ends to find:

#x = -6+-11sqrt(11)#

#color(white)()#
Bonus

It is very useful to know the first few powers of #11# ...

#11^0 = 1#

#11^1 = 11#

#11^2 = 121#

#11^3 = 1331#

#11^4 = 14641#

Compare this with:

#(a+b)^0 = 1#

#(a+b)^1 = a+b#

#(a+b)^2 = a^2+2ab+b^2#

#(a+b)^3 = a^3+3a^2b+3ab^2+b^3#

#(a+b)^4 = a^4+4a^3b+6a^2b^2+4ab^3+b^4#

Notice that the coefficients match the digits of the powers of #11#.

This is no coincidence: Let #a=10# and #b=1# to find out why.