How many zeroes does the function have #.25x^2-7x-12=y-4#?

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Answer 1 · 2015-08-07T00:36:13

All real quadratic functions have at most two zeroes.

For this one, if you multiply by whole thing by #4# and set #y = 0#:

#x^2 - 28x - 48 = -16#

#0 = x^2 - 28x - 32#

#x = (-(-28) pm sqrt((-28)^2 - 4(1)(-32)))/(2(1))#

#= (28 pm sqrt(784 + 128))/2#

#= (28 pm sqrt(912))/2 = (28 pm sqrt(57*4*4))/2#

#= (28 pm 4sqrt57)/2#

#= 14 pm 2sqrt57#

Thus the two roots are:

#color(blue)(x = 14 + 2sqrt57)#
#color(blue)(x = 14 - 2sqrt57)#