How long does it take the giraffe to weigh 85 kg?

  • Thread starter Thread starter Paulo2014
  • Start date Start date
Click For Summary

Homework Help Overview

The problem involves determining the time it takes for a giraffe to reach a weight of 85 kg, based on a quadratic equation that models its weight over the first two years of life. The equation provided is W=(t^2)/4-t+68, where W is the weight in kilograms and t is the time in months since birth.

Discussion Character

  • Mathematical reasoning, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • Participants discuss rearranging the quadratic equation to solve for t, with some expressing confusion about handling the fractional term. There are suggestions to eliminate fractions and to set the equation equal to zero for solving. Various methods for solving the quadratic equation are explored, including factoring and completing the square.

Discussion Status

The discussion is active, with participants providing guidance on manipulating the equation and exploring different solving techniques. There is recognition of the need to set the equation to zero before solving, and some participants express uncertainty about their approaches. Multiple interpretations of solving the equation are being explored without a clear consensus on the preferred method.

Contextual Notes

Participants mention the importance of reviewing textbook examples and concepts related to quadratic equations, indicating that some foundational knowledge may be lacking. There is also a note about the extraneous nature of negative solutions in the context of time.

Paulo2014
Messages
79
Reaction score
0

Homework Statement


The weight, W kg, of a giraffe over its first two years of life, is given by the equation
W=(t^2)/4-t+68
where t is the time in months since the giraffe was born.
How long does it take the giraffe to weigh 85 kg?

Homework Equations


The Attempt at a Solution

I know i need to rearage the equation but i can't remember how so any help is appreciated...
 
Last edited by a moderator:
Physics news on Phys.org
Which letter represents the weight? Put 85 there and solve it.
 
Yeah I understand that but how would I solve it?

its the (t^2)/4 bit i don't understand how to solve...
 
Let me see if this is the thing you have:
[tex]\frac{t^2}{4}-t+68=85[/tex]
That's a quadratic equation. Before you can solve, what does the equation have to look like?
Then, since fractions give me a headache, like like to ditch them. Can you think of a method of getting rid of them?
 
multiply 4 to both sides?
 
Yes please.
 
So it would be:
t^2-t+68=340?
 
so you have:
[tex]4(\frac{t^2}{4}-t+68=85)[/tex]
make sure that you are careful about your distributive property.
EVERYTHING gets multiplied by 4.
 
but doesn't the t^2 already have the /4 so wouldn't it be:

t^2+4(-t+68=85)?
 
  • #10
yes
[tex]4(\frac{t^2}{4}-t+68=85)[/tex]
[tex]t^2-(4)t+(4)68=(4)85[/tex]
then what?
 
  • #11
expand.

So it is t^2-4t+272=340

Is that right?
 
  • #12
yes.
In order to solve ANY equation, what does it have to be set equal to?
There's ONE MORE STEP before you can solve it.
 
  • #13
I can't remember... :(
 
  • #14
Look for an example in your book.
 
  • #15
It has to be in the form ax=b

thats what my textbook says... is that what you meant?
 
  • #16
You are looking for the ZEROS
What would you set the equation equal to to get the ZEROS?
 
  • #17
i'm lost...
 
  • #18
In order to solve any equation, it must be set equal to ZERO!
HOORAY!
How do you manipulate your equation to get it equal to zero?
 
  • #19
I see. take 340 from both sides so the equation is:t^2-4t-68=0 is that correct?
 
  • #20
Sit down and read the chapter ... do every single example problem! It will serve you better than trying to solve this problem.
 
  • #21
rocomath is right.
and you did get the equation right. If you aren't sure how to solve it, you DO need to read the chapter.
 
  • #22
I use the quadratic formula don't I...
 
  • #23
yes you do
 
  • #24
I got x=10.48528137423857 or -6.4852813742385695
 
  • #25
Is that right?
 
  • #27
t has to equal 10.5 as you can't have a negative amount of years right?
 
  • #28
Yes, the negative answer is extraneous.
 
  • #29
No, that's not what he is talking about.

happyg1, you don't have to set an expression to 0 to solve an equation. Here, of course, you could subtract 340 from both sides to get x2- 4t- 68= 0 and try to factor. However, I don't think it will be so easy to factor this.

My favored way of solving x2-4t+ 272=340 would be to subtract 272 from both sides: x2- 4t= 68 and then complete the square.
 
  • #30
So is my answer incorrect?
 

Similar threads

Replies
7
Views
2K
  • · Replies 10 ·
Replies
10
Views
6K
  • · Replies 15 ·
Replies
15
Views
5K
  • · Replies 18 ·
Replies
18
Views
3K
  • · Replies 7 ·
Replies
7
Views
6K
  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 6 ·
Replies
6
Views
4K
Replies
5
Views
7K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 2 ·
Replies
2
Views
3K