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

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