How do you solve #x^2-12x+32=0#?

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Answer 1 · 2016-06-22T18:03:36

#x=4# or #x=8#

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Method 1 - Finding a pair of factors

Find a pair of factors of #32# with sum #12#.

The pair #4, 8# works in that #4xx8 = 32# and #4+8=12#

Hence we find:

#0 = x^2-12x+32 = (x-4)(x-8)#

So #x = 4# or #x = 8#

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Method 2 - Completing the square

#0 = x^2-12x+32#

#= x^2-2(6x)+36-4#

#= (x-6)^2-2^2#

#= ((x-6)-2)((x-6)+2)#

#= (x-8)(x-4)#

Hence #x=8# or #x=4#

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Method 3 - Quadratic formula

#x^2-12x+32 = 0#

is in the form #ax^2+bx+c = 0# with #a=1#, #b=-12#, #c=32#

This has roots given by the quadratic formula:

#x = (-b+-sqrt(b^2-4ac))/(2a)#

#= (12+-sqrt(12^2-4(1)(32)))/(2*1)#

#= (12+-sqrt(144-128))/2#

#= (12+-sqrt(16))/2#

#= (12+-4)/2#

#= 6+-2#

So #x=8# or #x=4#