How do you solve #n^2-12n+27=0#?

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Answer 1 · 2016-08-21T09:47:06

#b = 3, b = 9#

#(n + b_1)(n + b_2) = n^2 + (b_1 + b_2)n + b_1b_2#

#n^2 - 12n + 27#

#=> b_1 + b_2 = -12#

#=> b_1b_2 = 27#

If we list down all factors of 27
#1*27#
#3*9#
#-1*-27#
#-3*-9#

We want a pair that should give us -12 when added.
Among the factors, #-3*-9# satisfy that condition

#n^2 - 12n + 27 = (b - 3)(b - 9) = 0#

Remember that for a product to be 0, either factor should be 0.
Which gives us

#b - 3 = 0 => b = 3#

#b - 9 = 0 => b = 9#

Answer 2 · 2016-08-21T09:47:42

The Soln.# : n=9, or, n=3#.

To solve #n^2-12n+27=0#, we have to factor the quadr. poly.

We observe that, #9xx3-=27, &, 9+3=12#. So, splitting #12n=9n+3n#, we have,

# n^2-12n+27=0#.

#rArr ul(n^2-9n)-ul(3n+27)=0#.

#rArr n(n-9)-3(n-9)=0#.

#rArr (n-9)(n-3)=0#.

#rArr n=9, or, n=3#.

These roots satisfy the gven eqn.

Hence, the Soln. # : n=9, or, n=3#.