# Quadratic Formula Question #5

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• May 6th 2013, 09:21 AM
Mathnood768
Quadratic Formula Question #5
http://i.imgur.com/TeW1TELh.jpg

I'm not sure how to do these word questions...

My attempt:

150x + 200x - 600 = 0

I suck at this, any help would be great! Thanks!
• May 6th 2013, 10:35 AM
Qwob
Re: Quadratic Formula Question #5
Think as travelling north as travelling in the y-plane, and travelling west as in the x-plane.

$\displaystyle speed = \frac {distance}{time}$

(or more correctly)

$\displaystyle velocity = \frac{displacement}{time}$

After 2 hours, the first plane will have travelled 300km. (150x2)

We can model this as y = 300 + 150t, where t is the time in hours

After 2 hours, second plane will have just taken off and therefore travelled 0km.

We can model this as x = 200t, where t is the time in hours.

Now, think of this as a vector triangle question like in physics. This way we have a right-angled triangle, and we want the hypotenuse (the distance between the two planes) to be 600km.

Therefore we have:

$\displaystyle x^2 + y^2 = 600^2$

$\displaystyle (200t)^2 + (300 + 150t)^2 = 600^2$

Now we solve for $\displaystyle t$:

$\displaystyle 40000t^2 + 90000 + 90000t + 22500t^2 = 360000$

simplifying:

$\displaystyle 400t^2 + 900 + 900t + 225t^2 = 3600$

$\displaystyle 80t^2 + 180 + 180t + 45t^2 = 720$

$\displaystyle 125t^2 + 180t - 540 = 0$

$\displaystyle 25t^2 + 36t - 108 = 0$

Using a calculator to quickly solve the quadratic:

t = 1.48 hours or t = -2.92 hours

since t cannot be negative, t must be 1.48 hours (or 1.5 hours to the nearest tenth of an hour).

This is just a solution to the problem. There may (and hopefully should) be a much easier way to solve it, maybe given as an example somewhere in your book. If not, I hope this helps.