How do you solve x^2-3x+2=0x2−3x+2=0?

1 Answer
George C.
Jun 25, 2015

0 = x^2-3x+2 = (x-1)(x-2)0=x2−3x+2=(x−1)(x−2)

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

Explanation:

Let f(x) = x^2-3x+2f(x)=x2−3x+2

First notice that since 1 - 3 + 2 = 01−3+2=0, we have f(1) = 0f(1)=0, x=1x=1 is a solution and (x-1)(x−1) is a factor of f(x)f(x).

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

(x-1)(x-2) = x^2 - 3x +2(x−1)(x−2)=x2−3x+2

... look at the coefficient of x^2x2 which comes from x*xx⋅x and the constant term 22, which comes from -1 xx -2−1×−2

So the other solution is x=2x=2.

Check: f(2) = 2^2-(3*2)+2 = 4 - 6 + 2 = 0f(2)=22−(3⋅2)+2=4−6+2=0

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