How do you solve #1/3t^2 + 3= 2t# using the formula?

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Answer 1 · 2015-06-27T20:24:28

First subtract #2t# from both sides to get:

#1/3t^2-2t+3 = 0#

#t = (2+-sqrt(2^2-(4xx1/3xx3)))/(2*1/3) = 3#

#1/3t^2-2t+3# is in the form #at^2+bt+c# with #a=1/3#, #b=-2# and #c=3#

So the roots of #1/3t^2-2t+3 = 0# are given by the quadratic formula:

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

#=(2+-sqrt(2^2-(4xx1/3xx3)))/(2*1/3)#

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

#=cancel(2)/(cancel(2)*1/3)#

#=3#

Alternatively, multiply the quadratic equation by #3# to get:

#t^2-6t+9 = 0#

Notice that the coefficients (ignoring signs) are 1,6,9.

Does the pattern #1, 6, 9# ring any bells?

Well #169 = 13^2# and non-coincidentally:

#t^2-6t+9 = (t-3)^2#

#t^2-6t+9 = 0# when #(t-3) = 0#, that is when #t=3#