How do you solve #T^2 + 7T - 2 = 0# by completing the square?

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Answer 1 · 2016-04-04T10:35:33

#T=-7/2+sqrt57/2# or #T=-7/2-sqrt57/2#

As #(x+a)^2=x^2+2ax+a^2#, to complete let us say

#x^2+2ax# to make complete square, we should add #a^2# or square of half of the coefficient of x.

Hence to solve #T^2+7T-2=0# we should add and subtract #(7/2)^2=49/4#.

Hence, #T^2+7T-2=0# can be written as #T^2+7T+49/4-49/4-2=0# or

#T^2+7T+49/4=49/4+2=57/4# or

= #(T+7/2)^2=57/4#

or #T+7/2=sqrt57/2# or #T+7/2=-sqrt57/2#

or #T=-7/2+sqrt57/2# or #T=-7/2-sqrt57/2#