First, find where the curve crosses the x-axis by factoring: y = 2x² - 3x - 4 = (2x + 1)(x - 2). So the points P and Q are at (-1/2,0) and (2,0).

Next, we need to know the slope of the curve at the points P and Q to construct tangent lines there. y' = 4x - 3. So the slope at P is -5 and the slope at Q is 5.

Now we know the equations of the tangent lines because we have the slope and a point that they pass through, P and Q.

y - 0 = -5(x - -1/2) --> y = -5x - 5/2

y - 0 = 5(x - 2) --> y = 5x - 10

We need to know where these lines intersect, which is point R. So set them equal to each other:

-5x - 5/2 = 5x - 10

10x = 10 - 5/2 = 15/2

**x = 15/20 = 3/4**

Plugging x back into one of the equations for the lines, we find that y = -25/4. Now we know that the point R is (3/4, -25/4)

Stepping back, we need to construct the normal lines to the tangent. This is easy, remembering that if the slope of the tangent line to P is m, then the slope of the normal line at the same point is -1/m. So repeat what we did above with the points P and Q to find their intersection at the point S.

y - 0 = (1/5)(x - -1/2) --> y = x/5 + 1/10

y - 0 = (-1/5)(x - 2) --> y = -x/5 + 2/5

x/5 + 1/10 = -x/5 + 2/5

2x/5 = 2/5 - 1/10 = 3/10

**x = 15/20 = 3/4**

--> y = (3/4)/5 + 1/10 = 3/20 + 1/10 = 5/20 = 1/4

So the point S is at (3/4, 1/4)

Now that we know the coordinates of the points R and S, we can find the distance between them using the equation d = sqrt( (x1 - x2)^2 + (y1 - y2)^2 )

d = sqrt( (3/4 - 3/4)^2 + (1/4 - -25/4)^2 )

d = 1/4 + 25/4 = 26/4 = 13/2 (note that the ^2 and sqrt cancel)

**d = 13/2**

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