Quadratic Equations

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stuffedfigs

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Oct 11, 2009
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Hey guys,

I've been stuck on these questions for a while now, and i've only been able to make "headway" with the first one, any help you could give you get me started would be greatly appreciated.

1. The height of a particular golf ball is given by the formula

h=30t-5t[sup:5zv6q0w8]2[/sup:5zv6q0w8]

where h is its height in metres and t is the time in seconds.

a) Write the formula in completed square form. I got 5(t-6)[sup:5zv6q0w8]2[/sup:5zv6q0w8]+30t-180=h. Anywhere near the answer?
b) Hence find the maximum height of the ball (Got to make t=0???)
c) After how many seconds does the ball reach its maximum height?

2. The path round a square lawn is 1 metre wide and made from rectangular slabs. If the area of the path and the area of the lawn are equal, find the area of the lawn, to 2 d.p. (Is that using the quadratic formula?)

3. A rectangle has width 1 unit and length x units. A square is cut off it. The rectangle left behind is similar to the original rectangle. Find the value of x in surd form. (Completely clueless to this one I'm afraid.)

As i said, any help would be greatly appreciated!!!
 
stuffedfigs said:
Hey guys,

I've been stuck on these questions for a while now, and i've only been able to make "headway" with the first one, any help you could give you get me started would be greatly appreciated.

1. The height of a particular golf ball is given by the formula

h=30t-5t[sup:2no18vw3]2[/sup:2no18vw3]

where h is its height in metres and t is the time in seconds.

a) Write the formula in completed square form. I got 5(t-6)[sup:2no18vw3]2[/sup:2no18vw3]+30t-180=h. Anywhere near the answer?<-----well, no, it isn't.
b) Hence find the maximum height of the ball (Got to make t=0???)
c) After how many seconds does the ball reach its maximum height?

2. The path round a square lawn is 1 metre wide and made from rectangular slabs. If the area of the path and the area of the lawn are equal, find the area of the lawn, to 2 d.p. (Is that using the quadratic formula?)

3. A rectangle has width 1 unit and length x units. A square is cut off it. The rectangle left behind is similar to the original rectangle. Find the value of x in surd form. (Completely clueless to this one I'm afraid.)

As i said, any help would be greatly appreciated!!!
Click to expand...

I'll do a problem SIMILAR to your first one.

Suppose the height of a ball is given by

h = 24t - 6t[sup:2no18vw3]2[/sup:2no18vw3]
where h is the height in metres and t is the time in seconds.

Write this in "completed square" form.

Start by removing a factor of -6 from both terms of the right side (so that you'll end up with a coefficient of +1 for t[sup:2no18vw3]2[/sup:2no18vw3]:

h = -6 (t[sup:2no18vw3]2[/sup:2no18vw3] - 4t + ???)

Now, we'll replace the ??? with a number that will make what's inside the parentheses a perfect square. Take half the coefficient of t, square that, and that's what will make the expression a perfect square. Half of -4 is -2, and (-2)[sup:2no18vw3]2[/sup:2no18vw3] is 4. Replace the ??? with 4....but because of the factor of -6 sitting outside the parentheses, we've really added -24 to the right side. To balance that out, add + 24 to that side as well.

h = -6 (t[sup:2no18vw3]2[/sup:2no18vw3] - 4t + 4) + 24

or,

h = -6(t - 2)[sup:2no18vw3]2[/sup:2no18vw3] + 24

Do you recognize this as the "vertex form" of the equation of a parabola? When the equation of a parabola is in the form
y = a(x - h)[sup:2no18vw3]2[/sup:2no18vw3] + k, the vertex (maximum or minimum point for the parabola) is at (h, k). If "a" is positive, the parabola opens upward and has a minimum point. If "a" is negative, the parabola opens downward and has a maximum.

You've got

h = -6 (t - 2)[sup:2no18vw3]2[/sup:2no18vw3] + 24

a = -6, so the parabola opens downward. The vertex is at (2, 24), and it is a maximum. So.....what does this tell you? Well, at t = 2 seconds, the height h = 24 metres. You know the maximum height the ball reaches now, and you know how many seconds it will take to reach that maximum height.

Try similar steps on your first problem.
 
Aha!

Many thanks for the reply!!

I now got the answer to be h=-5(t-3)[sup:2ftqbom8]2[/sup:2ftqbom8]+45

So the Height = 45 metres and the time to get there is 3 seconds.
 
Blondiemathematician said:
Just wondering if anyone actually solved questions 2 & 3 as I'm doing the exact ones and still incredibly stuck... thanks
Click to expand...
Start a thread of your own, Blondie.
 
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