The line $y = 3$ intersects the graph of $y = 4x^2 + x - 1$ at the points $A$ and $B$. The distance between $A$ and $B$ can be written as $\frac{\sqrt{m}}{n}$, where $m$ and $n$ are positive integers that do not share any factors other than one. Find the value of $m - n$.

Respuesta :

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].

gmany

Answer:

m - n = 61

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]

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