If the lines $$x + y = a$$ and $$x - y = b$$ touch the curve $$y = x^2 - 3x + 2$$ at the points where the curve intersects the $$x$$-axis, then $$\frac{a}{b}$$ is equal to...

We begin with the curve $$y = x^2 - 3x + 2$$. The points where this curve meets the $$x$$-axis are obtained by setting $$y = 0$$ because every $$x$$-axis point has ordinate zero.

So we solve $$x^2 - 3x + 2 = 0$$. Factoring, we have$$(x - 1)(x - 2) = 0,$$which gives the two roots$$x = 1 \quad\text{and}\quad x = 2.$$

Hence the curve intersects the $$x$$-axis at the points$$(1,\,0)\quad\text{and}\quad(2,\,0).$$

According to the question, the straight line $$x + y = a$$ touches (is tangent to) the curve at one of these points, and the straight line $$x - y = b$$ touches the curve at the other. For a line to be a tangent at a point on a curve, its slope must equal the derivative of the curve at that point.

The derivative of the curve is obtained using the power rule $$\frac{d}{dx}\,(x^n)=nx^{\,n-1},$$so

$$\frac{dy}{dx}=2x-3.$$

Tangent at the point $$(1,\,0)$$

Substituting $$x = 1$$ into the derivative gives the slope at this point:$$m_1 = 2(1) - 3 = -1.$$

Now we write the tangent line through $$(1,\,0)$$ with slope $$-1$$ using the point-slope form $$y - y_1 = m(x - x_1):$$ $$y - 0 = -1\bigl(x - 1\bigr).$$ Simplifying, we get$$y = -x + 1.$$ Rearranging into the standard form $$x + y = \text{constant},$$ we have$$x + y = 1.$$ Thus$$a = 1.$$

Tangent at the point $$(2,\,0)$$

Substituting $$x = 2$$ into the derivative gives the slope at this point:$$m_2 = 2(2) - 3 = 1.$$

Again using the point-slope form through $$(2,\,0)$$:$$y - 0 = 1\bigl(x - 2\bigr).$$ Hence$$y = x - 2.$$ Rearranging into the form $$x - y = \text{constant},$$ we obtain$$x - y = 2.$$ Therefore$$b = 2.$$

We now compute the required ratio: $$\frac{a}{b} = \frac{1}{2} = 0.5.$$

So, the answer is $$0.5$$.