Expert: Paul Klarreich - 7/17/2007

Question
Hi I have a few questions I need help understanding hopefully you can help?

1. Solve by completing the square I have the answer
x^2-7x-2=0   the answer I came up with is (7+/- sqrt of 57/2

2. (2x-5)(x+1)=2

3. 9-6x+x^2=0

4. (x^2/x+6)times (x^2+x-30/x^2-2x)

5. divide ((3x^2+14x+15)/(3x^2+8x+5))/((x^2-x-12)/(-3x^2+9x+12))

6 This is the word problem I understand the formula but how can I get this with no information.
A rock is thrown from the top of a building. The distance, in feet, between the rock and the ground t seconds after the rock is thrown is given by d = -16t^2 - 6t + 27. How long after the rock is thrown does it hit the ground?


Answer
Questioner:   Wade
Category:  Advanced Math
Private:  No
 
Subject:  Math questions
Question:  Hi I have a few questions I need help understanding hopefully you can help?

1. Solve by completing the square I have the answer
x^2-7x-2=0   the answer I came up with is (7+/- sqrt of 57/2

2. (2x-5)(x+1)=2

3. 9-6x+x^2=0

4. (x^2/x+6)times (x^2+x-30/x^2-2x)

5. divide ((3x^2+14x+15)/(3x^2+8x+5))/((x^2-x-12)/(-3x^2+9x+12))

6 This is the word problem I understand the formula but how can I get this with no information.
A rock is thrown from the top of a building. The distance, in feet, between the rock and the ground t seconds after the rock is thrown is given by d = -16t^2 - 6t + 27. How long after the rock is thrown does it hit the ground?
...........................................................
Hi, Wade,

A few questions, he says. OK, I'll see what I can do.  


1. Completing the square:

x^2 - 7x - 2 =  0

x^2 - 7x     =  2

x^2 - 7x + 49/4  =  2 + 49/4

  (x - 7/2)^2   =  57/4

  x - 7/2 = +- sqrt(57)/2
        x = 7/2 +- sqrt(57)/2

Looks good.
.........................
2. (2x-5)(x+1)=2

Multiply out, simplify, solve:

2x^2 - 3x - 5 = 2

2x^2 - 3x - 7 = 0

I think you'll use the quadratic formula now.
...........................

3. 9-6x+x^2=0

Rewrite:

x^2 - 6x + 9 = 0

(x - 3)(x - 3) = 0

x = 3 is a double root.

................................

4. (x^2/x+6)times (x^2+x-30/x^2-2x)
x^2    x^2 + x - 30
-----  -------------
x + 6     x^2 - 2x
Factor:

x x    (x + 6)(x - 5)
-----  ---------------
x + 6     x(x - 2)

Cancel factors:

x .    .......(x - 5)
-----  ---------------
.....     .(x - 2)

Simplify:

x(x - 5)
--------
x - 2
.........................

5. divide ((3x^2+14x+15)/(3x^2+8x+5))/((x^2-x-12)/(-3x^2+9x+12))

3x^2 + 14x +15   -3x^2 + 9x + 12
---------------  ---------------   << Note the inversion.
3x^2 + 8x + 5     x^2 - x - 12

Factor all the numerators and denominators.  Note the sneaky factoring trick at the end.

3x^2 + 14x +15   -3(x^2 - 3x - 4)
---------------  ----------------
3x^2 + 8x + 5     (x - 4)(x + 3)

(x + 3)(3x + 5)  -3(x - 4)(x + 1)
---------------  ----------------
(x + 1)(3x + 5)   (x - 4)(x + 3)

Cancel common factors: (I think you can handle that.)

Sneaky factoring trick:  3x^2 + 14x + 15  and 3x^2 + 8x + 5  are nasty factorings, so leave them for last.  Then you see what other factors have come up and you try those first.
.......................................
6 This is the word problem I understand the formula but how can I get this with no information.
A rock is thrown from the top of a building. The distance, in feet, between the rock and the ground t seconds after the rock is thrown is given by d = -16t^2 - 6t + 27. How long after the rock is thrown does it hit the ground?

I am sorry, but to get the formula you will need some calculus, which is in the future.  

Meanwhile:

d(t) = -16t^2 - 6t + 27

This encapsulates three constants:

-16 is  1/2 the acceleration due to gravity, a constant on the Earth's surface.
-6 is the initial speed, indicating it was thrown DOWNWARD at 6 ft/sec.
27 is the height of the building.

The rock hits the ground when  d(t) = 0, so solve:

-16t^2 - 6t + 27 = 0

16t^2 + 6t - 27 = 0
(8t - 9)(2t + 3) = 0

t = 8/9  (good)
t = -3/2  (not good; before the throwing)