Answer: 5
Step-by-step explanation:
First, find the intersections of the parabolas using any method. I choose to use the elimination method
[tex]y=x^2-\dfrac{11}{4}x-\dfrac{7}{4}\ \rightarrow \ -8\bigg(y=x^2-\dfrac{11}{4}x-\dfrac{7}{4}\bigg)\ \rightarrow \ -8y=-8x^2+22x+14\\\\\\y=-\dfrac{7}{8}x^2+x+\dfrac{31}{8}\ \rightarrow \ 8\bigg(y=-\dfrac{7}{8}x^2+x+\dfrac{31}{8}\bigg)\ \rightarrow \underline{\ 8y=-7x^2+\ 8x+31\ }\\\\.\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad\qquad \qquad \qquad 0=-15x^2+30x+45[/tex]
[tex]\text{Solve the quadratic equation:}\\0=-15x^2+30x+45\\.\ =-15(x^2-2x-3)\\.\ =-15(x-3)(x+1)\\\\0=x-3\quad and\quad 0=x+1\\x=3\qquad and \qquad x=-1\\\\\text{Plug in the x-values into either equation to find the corresponding y-values:}\\y=(3)^2-\dfrac{11}{4}(3)-\dfrac{7}{4}\\\\.\ =9-\dfrac{33}{4}-\dfrac{7}{4}\\\\.\ =\dfrac{36}{4}-\dfrac{33}{4}-\dfrac{7}{4}\\\\.\ =-\dfrac{4}{4}\\\\.\ =-1\\\\\text{The coordinate of one of the intersections is:}\ (3, -1)[/tex]
[tex]y=(-1)^2-\dfrac{11}{4}(-1)-\dfrac{7}{4}\\\\.\ =1+\dfrac{11}{4}-\dfrac{7}{4}\\\\.\ =\dfrac{4}{4}+\dfrac{11}{4}-\dfrac{7}{4}\\\\.\ =\dfrac{8}{4}\\\\.\ =2\\\\\text{The coordinate of the other intersection is:}\ (-1, 2)[/tex]
Next, find the distance between the intersected points (3, -1) & (-1, 2):
[tex]d=\sqrt{(3+1)^2+(-1-2)^2}\\.\ =\sqrt{(4)^2+(-3)^2}\\.\ =\sqrt{16+9}\\.\ =\sqrt{25}\\.\ =5[/tex]
