#1
Hi all,

Is there someone who can help me solving this question. I do not understand what they mean by saying, that a quadratic formule has two answer a and b. I really do'nt get it.

24 A quadratic equation has solutions a and b.
a + b = -5 ab = 3
What is the equation?
A x2 + 5 x + 3 = 0
B x2 - 5 x - 3 = 0
C x2 - 5 x + 3 = 0
D x2 + 3 x + 5 = 0
E x2 + 5 x - 3 = 0
F x2 + 3 x - 5 = 0
0
4 years ago
#2
(Original post by Roqia)
Hi all,

Is there someone who can help me solving this question. I do not understand what they mean by saying, that a quadratic formula has two answer a and b. I really do'nt get it.

What work have you done with quadratics before?
0
4 years ago
#3
if a and b are roots, then it can be factorised:

0
#4
(Original post by Hasufel)
if a and b are roots, then it can be factorised:

Great!! I got it!! THanks a lot

27 Taking the speed of sound in air as 300 m/s and in steel as 4800 m/s. A worker lies next to a
railway line and hears the whistle of a train through the steel rails and 1.5 s later hears the
same whistle through the air.
How far away was the train when its whistle sounded?
A 200 m
B 450 m
C 480 m
D 4500 m
E 6750 m
F 7200 m
0
4 years ago
#5
(Original post by Roqia)
Great!! I got it!! THanks a lot

27 Taking the speed of sound in air as 300 m/s and in steel as 4800 m/s. A worker lies next to a
railway line and hears the whistle of a train through the steel rails and 1.5 s later hears the
same whistle through the air.
How far away was the train when its whistle sounded?
A 200 m
B 450 m
C 480 m
D 4500 m
E 6750 m
F 7200 m
unfortunately, (no offence) questions like this repulse me - since i hated them at university, and had my fill of mechanics!

Anyone else want to help?
0
4 years ago
#6
(Original post by Roqia)

Taking the speed of sound in air as 300 m/s and in steel as 4800 m/s. A worker lies next to a railway line and hears the whistle of a train through the steel rails and 1.5 s later hears the
same whistle through the air.
How far away was the train when its whistle sounded?
What have you tried?

If you set the distance as d you should be able to get two expressions for time and then use the difference of 15 to form an equation
0
#7
(Original post by Hasufel)
unfortunately, (no offence) questions like this repulse me - since i hated them at university, and had my fill of mechanics!

Anyone else want to help?
No problem, thanks anyway Its a very tricky one, and I just don't get it Been trying 15 minutes now ..
0
4 years ago
#8
(Original post by Roqia)
No problem, thanks anyway Its a very tricky one, and I just don't get it Been trying 15 minutes now ..
Actually it is pretty straightforward - have you tried the approach I suggested?
0
#9
(Original post by TenOfThem)
Actually it is pretty straightforward - have you tried the approach I suggested?
Hello, thanks fro your comment !

This is what I tried (4800 x d) - (300 x d) = 1.5

d= 3000 but this is not the correct answer
0
4 years ago
#10
(Original post by Roqia)
Hello, thanks fro your comment !

This is what I tried (4800 x d) - (300 x d) = 1.5

d= 3000 but this is not the correct answer
Do you know how to find Time when you have Distance and Speed?
0
#11
(Original post by TenOfThem)
Do you know how to find Time when you have Distance and Speed?

I guess it should be this then (d/4800) - (d/300) = 1.5

I got the right answer now if i solve this !!!

THANKS!!!
0
4 years ago
#12
(Original post by Roqia)

I guess it should be this then (d/4800) - (d/300) = 1.5

I got the right answer now if i solve this !!!

THANKS!!!
np
0
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