How do you solve #t^ { 2} + 4t - 12= 0#?

1 Answer
George C.
Feb 18, 2017

#t = -6" "# or #" "t = 2#

Explanation:

Here are a couple of methods...

#color(white)()#
Method 1 - Fishing for factors

Find a pair of factors of #12# which differ by #4#.

The pair #6, 2# works and hence we find:

#0 = t^2+4t-12 = (t+6)(t-2)#

So:

#t = -6" "# or #" "t = 2#

#color(white)()#
Method 2 - Completing the square

The difference of squares identity can be written:

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

Complete the square and use this with #a=(t+2)# and #b=4# as follows:

#0 = t^2+4t-12#

#color(white)(0) = t^2+4t+4-16#

#color(white)(0) = t^2+2(t)(2)+2^2-16#

#color(white)(0) = (t+2)^2-4^2#

#color(white)(0) = ((t+2)-4)((t+2)+4)#

#color(white)(0) = (t-2)(t+6)#

So:

#t = 2" "# or #" "t = -6#

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