The Student Room Group
Uni students - Complete our survey >>
Uni students - Complete our survey >>
Uni students - Complete our survey >>

Maths Question Help

Guys, I need help on this question. So far I've got:
4.9t^2 -13.2t+6<0
4.9t^2-13t+6.3<0
(edited 2 years ago)
Q10_2.4-Inequalities_Very_Hard_A_Level_Maths_Pure-1.png20.2KB
Reply 1
Avatar for mqb2766
mqb2766
21
Think you need to check your constant terms.
Reply 2
Avatar for Nithu05
Nithu05
OP
15
Original post by mqb2766
Think you need to check your constant terms.

Well, they're both projected from different heights so I added 4 onto the initial heights but now that I think about it, I think you need to subtract 4 from the heights? However, even if I do this I'll end up with two quadratics. How am I supposed to solve them simultaneously? Am I supposed to work out the inequalities separately and then combine them?
Reply 3
Avatar for mqb2766
mqb2766
21
Original post by Nithu05
Well, they're both projected from different heights so I added 4 onto the initial heights but now that I think about it, I think you need to subtract 4 from the heights? However, even if I do this I'll end up with two quadratics. How am I supposed to solve them simultaneously? Am I supposed to work out the inequalities separately and then combine them?

Rather than guessing what to do, write each quadratic inequality out fully, then collect terms.
As a fall back working them out seperately then combining will give a solution. It is only a couple of quadratics, should just take a couple of mins.
(edited 2 years ago)
Reply 4
Avatar for Nithu05
Nithu05
OP
15
Original post by mqb2766
Rather than guessing what to do, write each quadratic inequality out fully, then collect terms.
As a fall back working them out seperately then combining will give a solution. It is only a couple of quadratics, should just take a couple of mins.

Ok, so far I got:
4.9t^2-13t+1.7<=0
4.9t^2-13.2t+2<=0
Reply 5
Avatar for Nithu05
Nithu05
OP
15
Original post by Nithu05
Ok, so far I got:
4.9t^2-13t+1.7<=0
4.9t^2-13.2t+2<=0

Thing is when I'm working out the roots, they're not whole numbers so something seems wrong.
Reply 6
Avatar for mqb2766
mqb2766
21
Original post by Nithu05
Ok, so far I got:
4.9t^2-13t+1.7<=0
4.9t^2-13.2t+2<=0


that looks better, so just solve each of them.
Reply 7
Avatar for mqb2766
mqb2766
21
Original post by Nithu05
Thing is when I'm working out the roots, they're not whole numbers so something seems wrong.

It doesn't look like a particularly "nice" problem in that respect.
https://www.desmos.com/calculator/s5sbdhrsaa
The quadratics numbers are too close, as you can see from the graphs.

You can always verify by subbing the times back into the original quadratics.
(edited 2 years ago)
Reply 8
Avatar for Nithu05
Nithu05
OP
15
Original post by mqb2766
It doesn't look like a particularly "nice" problem in that respect.
https://www.desmos.com/calculator/s5sbdhrsaa
The quadratics numbers are too close, as you can see from the graphs.

You can always verify by subbing the times back into the original quadratics.

OK, I worked out the combined inequality and got:
0.1611559962<=t<=2.515119982
Reply 9
Avatar for Nithu05
Nithu05
OP
15
Original post by Nithu05
OK, I worked out the combined inequality and got:
0.1611559962<=t<=2.515119982

So is this the answer or is there one more step? Thanks so much for your help so far!
Original post by Nithu05
OK, I worked out the combined inequality and got:
0.1611559962<=t<=2.515119982

Possibly a bit too precise for quoting, but just sub back into the quadratics to verify if you want.
They look about right from desmos.
Reply 11
Avatar for Nithu05
Nithu05
OP
15
Original post by mqb2766
Possibly a bit too precise for quoting, but just sub back into the quadratics to verify if you want.
They look about right from desmos.

Ok I've substituted the values and it satisfies the inequalities. Should I just leave my answer as this?
Original post by Nithu05
Ok I've substituted the values and it satisfies the inequalities. Should I just leave my answer as this?

Probably not quite that precise (maybe 2 or 3 dp), but yes;
Reply 13
Avatar for Nithu05
Nithu05
OP
15
Original post by mqb2766
Probably not quite that precise (maybe 2 or 3 dp), but yes;

Ok thanks so much for your help :smile:

Quick Reply

Related discussions

Latest

Trending

Trending

Articles for you