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

Jun 25, 2015

$0 = {x}^{2} - 3 x + 2 = \left(x - 1\right) \left(x - 2\right)$

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

Explanation:

Let $f \left(x\right) = {x}^{2} - 3 x + 2$

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

We can see that the other factor must be $\left(x - 2\right)$ in order that:

$\left(x - 1\right) \left(x - 2\right) = {x}^{2} - 3 x + 2$

... look at the coefficient of ${x}^{2}$ which comes from $x \cdot x$ and the constant term $2$, which comes from $- 1 \times - 2$

So the other solution is $x = 2$.

Check: $f \left(2\right) = {2}^{2} - \left(3 \cdot 2\right) + 2 = 4 - 6 + 2 = 0$