Answer:
61
Step-by-step explanation:
Let's find the points [tex]A[/tex] and [tex]B[/tex].
We know that the [tex]y[/tex]-coordinates of both are [tex]3[/tex].
So let's first solve:
[tex]3=4x^2+x-1[/tex]
Subtract 3 on both sides:
[tex]0=4x^2+x-1-3[/tex]
Simplify:
[tex]0=4x^2+x-4[/tex]
I'm going to use the quadratic formula, [tex]x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}[/tex], to solve.
We must first compare to the quadratic equation, [tex]ax^2+bx+c=0[/tex].
[tex]a=4[/tex]
[tex]b=1[/tex]
[tex]c=-4[/tex]
[tex]\frac{-1 \pm \sqrt{1^2-4(4)(-4)}}{2(4)}[/tex]
[tex]\frac{-1 \pm \sqrt{1+64}}{8}[/tex]
[tex]\frac{-1 \pm \sqrt{65}}{8}[/tex]
Since the distance between the points [tex]A[/tex] and [tex]B[/tex] is horizontal. We know this because they share the same [tex]y-coordinate[/tex].This means we just need to find the positive difference between the [tex]x[/tex]-values we found for the points of [tex]A[/tex] and [tex]B[/tex].
So that is, the distance between [tex]A[/tex] and [tex]B[/tex] is:
[tex]\frac{-1+\sqrt{65}}{8}-\frac{-1-\sqrt{65}}{8}[/tex]
[tex]\frac{-1+\sqrt{65}+1+\sqrt{65}}{8}[/tex]
[tex]\frac{2\sqrt{65}}{8}[/tex]
[tex]\frac{\sqrt{65}}{4}[/tex]
If we compare this to [tex]\frac{\sqrt{m}}{n}[/tex], we should see that:
[tex]m=65 \text{ and } n=4[/tex].
So [tex]m-n=65-4=61[/tex].
Answer:
Step-by-step explanation:
[tex]y=4x^2+x-1\\\\y=3\\\\\text{Find the points A and B}.\\\\4x^2+x-1=3\qquad\text{subtract 3 from both sides}\\\\4x^2+x-1-3=3-3\\\\4x^2+x-4=0[/tex]
[tex]\text{Use the quadratic formula:}\\\\a=4,\ b=1,\ c=-4\\\\b^2-4ac=1^2-4(4)(-4)=1+64=65\\\\x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}\to x=\dfrac{-1\pm\sqrt{65}}{2(4)}=\dfrac{-1\pm\sqrt{65}}{8}[/tex]
[tex]\text{Therefore}\\\\A\left(\dfrac{-1-\sqrt{65}}{8},\ 3\right),\ B\left(\dfrac{-1+\sqrt{65}}{8},\ 3\right)\\\\\text{The formula of a distance between two points:}\\\\d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\\\\\text{Substitute:}\\\\d=\sqrt{\left(\dfrac{-1+\sqrt{65}}{8}-\dfrac{-1-\sqrt{65}}{8}\right)^2+(3-3)^2}\\\\d=\sqrt{\left(\dfrac{-1+\sqrt{65}-(-1)-(-\sqrt{65})}{8}\right)^2+0^2}\\\\d=\sqrt{\left(\dfrac{-1+\sqrt{65}+1+\sqrt{65}}{8}\right)^2}\\\\d=\sqrt{\left(\dfrac{2\sqrt{65}}{8}\right)^2}[/tex]
[tex]d=\sqrt{\left(\dfrac{\sqrt{65}}{4}\right)^2}\Rightarrow d=\dfrac{\sqrt{65}}{4}\Rightarrow\dfrac{\sqrt{65}}{4}=\dfrac{\sqrt{m}}{n}\\\\\text{Therefore}\ m=65\ \text{and}\ n=4\\\\m-n=65-4=61[/tex]
