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#1
Hi guys,

So I do have:
1.25t^2+8t-30=0

Then I am using quadratic formula as below:

x=-8-4*1.25*40/2*1.25

but the answers show me that I should do it like this:

x=-8+4*1.25*40/2*1.25

Why is there +4 and not -4?
What am I doing wrong?

0
6 months ago
#2
(Original post by Anna_anna_anna)
Hi guys,

So I do have:
1.25t^2+8t-30=0

Then I am using quadratic formula as below:

x=-8-4*1.25*40/2*1.25

but the answers show me that I should do it like this:

x=-8+4*1.25*40/2*1.25

Why is there +4 and not -4?
What am I doing wrong?

Should t be positive? What is the original question?

Generally there are two solutions so the +/- in the quadratic formula (outside the sqrt(), is your discriminant correct?), but depending on the question, only one may make sense.
Last edited by mqb2766; 6 months ago
0
#3
(Original post by mqb2766)
Should t be positive? What is the original question?

Generally there are two solutions so the +/- in the quadratic formula (outside the sqrt(), is your discriminant correct?), but depending on the question, only one may make sense.

O.K., I will try again as I think that I didn't explain that clearly...

I have to find out what is t, which stands for the time of the particle moving from A to B (there is only one answer and t should be positive):

So from:

1.25t^2+8t-30=0

I used quadratic formula as below:

x=-8+√8^2-4*1.25*40/2*1.25

but the answers show me that I should do it like this:

x=-8+√8^2+4*1.25*40/2*1.25

I thought that I always should use -4 as it is in the original formula...
Why is there +4 and not -4? When should I use +4?
What am I doing wrong?
0
6 months ago
#4
The -30 (or -40?) i.e. the constant term in the quadratic is negative so
-4ac = + 4*1.25*40
So two negatives (-4 and -40) are multiplied to produce a positive term.
Last edited by mqb2766; 6 months ago
1
#5
(Original post by mqb2766)
The -30 (or -40?) i.e. the constant term in the quadratic is negative so
-4ac = + 4*1.25*40
So two negatives (-4 and -40) are multiplied to produce a positive term.
Thank you 0
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