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How do you find all the zeros of #f(x) = 4(x + 3)^2 - 1#?

1 Answer

Mar 5, 2016

Set #f(x) = 0# and solve for #x# to find that #f(x)# has zeroes at #x=-7/2# and #x=-5/2#

Explanation:

#4(x+3)^2-1 = 0#

#=> 4(x+3)^2 = 1#

#=> (x+3)^2 = 1/4#

#=> x+3 = +-sqrt(1/4) = +-1/2#

#=>x = -3+-1/2#

#:. x = -7/2# or #x = -5/2#

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