How do you solve #2w ^ { 2} + 11w + 39= ( w + 7) ^ { 2}#?

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Answer 1 · 2017-08-31T06:57:02

#w = 5" "# or #" "w = -2#

Note that:

#(w+7)^2 = w^2+14w+49#

So the given equation can be rewritten as:

#2w^2+11w+39 = w^2+14w+49#

Subtracting #w^2+14w+49# from both sides, this becomes:

#w^2-3w-10 = 0#

We can factor this by noting that #5-2 = 3# and #5*2 = 10#, so:

#0 = w^2-3w-10 = (w-5)(w+2)#

Hence the solutions are:

#w = 5" "# or #" "w = -2#