How do you solve #x^2-3x+2=0#?

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Answer 1 · 2015-06-25T18:40:25

#0 = x^2-3x+2 = (x-1)(x-2)#

So #x=1# and #x=2# are the solutions.

Let #f(x) = x^2-3x+2#

First notice that since #1 - 3 + 2 = 0#, we have #f(1) = 0#, #x=1# is a solution and #(x-1)# is a factor of #f(x)#.

We can see that the other factor must be #(x-2)# in order that:

#(x-1)(x-2) = x^2 - 3x +2#

... look at the coefficient of #x^2# which comes from #x*x# and the constant term #2#, which comes from #-1 xx -2#

So the other solution is #x=2#.

Check: #f(2) = 2^2-(3*2)+2 = 4 - 6 + 2 = 0#