Question

What is the length of the segment, endpoints of which are intersections of parabolas y=x^2− 11/4 x− 7/4 and y=− 7 /8 x^2+x+ 31/8 ?

Answer 1

Answer: 5

Explanation:

First, find the intersections of the parabolas using any method. I choose to use the elimination method

`y=x^2-(11)/(4)x-(7)/(4)\ \rightarrow \ -8\bigg(y=x^2-(11)/(4)x-(7)/(4)\bigg)\ \rightarrow \ -8y=-8x^2+22x+14\\\\\\y=-(7)/(8)x^2+x+(31)/(8)\ \rightarrow \ 8\bigg(y=-(7)/(8)x^2+x+(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`

`\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-(11)/(4)(3)-(7)/(4)\\\\.\ =9-(33)/(4)-(7)/(4)\\\\.\ =(36)/(4)-(33)/(4)-(7)/(4)\\\\.\ =-(4)/(4)\\\\.\ =-1\\\\\text{The coordinate of one of the intersections is:}\ (3, -1)`

`y=(-1)^2-(11)/(4)(-1)-(7)/(4)\\\\.\ =1+(11)/(4)-(7)/(4)\\\\.\ =(4)/(4)+(11)/(4)-(7)/(4)\\\\.\ =(8)/(4)\\\\.\ =2\\\\\text{The coordinate of the other intersection is:}\ (-1, 2)`

Next, find the distance between the intersected points (3, -1) & (-1, 2):

`d=√((3+1)^2+(-1-2)^2)\\.\ =√((4)^2+(-3)^2)\\.\ =√(16+9)\\.\ =√(25)\\.\ =5`