How do you factor #w^2 + 14w-1632=0#?

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Answer 1 · 2016-07-19T11:22:02

#(w-34)(w+48) = 0#

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Method 1 - Completing the square

#0 = w^2+14w-1632#

#= (w+7)^2-49-1632#

#= (w+7)^2-1681#

#= (w+7)^2-41^2#

#= ((w+7)-41))((w+7)+41)#

#= (w-34)(w+48)#

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Method 2 - Find pair of factors

We want to find a pair of factors of #1632# which differ by #14#.

Since #14# is somewhat smaller than #1632# they will be approximately #sqrt(1632)-7# and #sqrt(1632)+7#.

Note that #40^2 = 1600# and #41^2 = 1681#

So:

#sqrt(1632) ~~ 41.5#

#sqrt(1632)-7 ~~ 34.5#

#sqrt(1632)+7 ~~ 48.5#

Of the nearby numbers, find that #34# and #48# are both factors of #1632# and they differ by #14#, so:

#w^2+14w-1632 = (w+48)(w-34)#