What is the smaller value of x for which x\(^2\) - 3x + 2= 0?

The given equation is a quadratic equation in standard form, which is \(ax^2 + bx + c = 0\), where \(a = 1\), \(b = -3\), and \(c = 2\). We can solve for the roots of the quadratic equation by using the quadratic formula: \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\] Substituting the values of \(a\), \(b\), and \(c\) into the formula, we get: \[x = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(1)(2)}}{2(1)}\] \[x = \frac{3 \pm \sqrt{1}}{2}\] Thus, the solutions are: \[x_1 = \frac{3 - 1}{2} = 1 \text{ and } x_2 = \frac{3 + 1}{2} = 2\] Therefore, the smaller value of \(x\) is \(1\). Hence, the answer is 1.