You are Here: Home >< Maths

# C1 Help ? watch

1. If anyone could give me a hand with this question and point out where i am going wrong and point me in the right direction it would be appreciated.

Find the range of values for p for which the followed equations have no real roots,

(2p + 1)x^2 - 3px + 2p = 4

a=(2p + 1)
b=3p
c=2p

b^2 -4(a)(c)

9p^2 - 4(2p + 1)(2) = 4

9p^2 - 16p - 8 = 4

9p^2 - 16p - 4 = 0

(3p - 4)^2 - 16 - 4 = 0
2. (Original post by Shakaa)
If anyone could give me a hand with this question and point out where i am going wrong and point me in the right direction it would be appreciated.

Find the range of values for p for which the followed equations have no real roots,

(2p + 1)x^2 - 3px + 2p = 4

a=(2p + 1)
b=3p
c=2p

b^2 -4(a)(c)

9p^2 - 4(2p + 1)(2) = 4

9p^2 - 16p - 8 = 4

9p^2 - 16p - 4 = 0

(3p - 4)^2 - 16 - 4 = 0

(2p + 1)x^2 - 3px + 2p = 4
(2p + 1)x^2 - 3px + 2p - 4 = 0

a = 2p + 1
b = -3p (dont forget the minus sign)
c = 2p - 4

and no real roots!! therefore b^2 - 4ac < 0
3. Firstly, get all your stuff in the LHS so RHS = 0

Then work out your b^2 - 4ac which is a little different because c differs.

Then remember that b^2 - 4ac must < 0 for no real roots

It's an inequality, not equation
4. (Original post by Shakaa)
If anyone could give me a hand with this question and point out where i am going wrong and point me in the right direction it would be appreciated.

Find the range of values for p for which the followed equations have no real roots,

(2p + 1)x^2 - 3px + 2p = 4

a=(2p + 1)
b=3p
c=2p

b^2 -4(a)(c)

9p^2 - 4(2p + 1)(2) = 4

9p^2 - 16p - 8 = 4

9p^2 - 16p - 4 = 0

(3p - 4)^2 - 16 - 4 = 0
does the bit i highlighted make sense?
5. (Original post by dexter -1)
does the bit i highlighted make sense?
It does dexter cheers.
6. Still having issues with this :s I have corrected all the mistakes i can find but i still cant get the correct values

(2p + 1)x^2 - 3px + 2p = 4

a=(2p + 1)
b=-3p
c=2p - 4

b^2 -4(a)(c)

9p^2 -4(2p + 1)(2p - 4)

9p^2 -16p^2 + 32p - 8p + 16

-7p^2 + 24p + 16 <0
(Is this right and if so where do i take it from here?)
7. You have to find the range(s) of p that satisfy the inequality.

You say this is a C1 question. Solving the inequality feels like an FP1 question, but hey.

You could graph it.

Or you could factorise the LHS into two linear factors (take the minus sign out first).

By thinking about when each factor is positive/negative you can solve the inequality
8. (Original post by ian.slater)
You have to find the range(s) of p that satisfy the inequality.

You say this is a C1 question. Solving the inequality feels like an FP1 question, but hey.

You could graph it.

Or you could factorise the LHS into two linear factors (take the minus sign out first).

By thinking about when each factor is positive/negative you can solve the inequality
I know its somewhat frowned up, but could you give me a example of how you would proceed with this question?
9. (Original post by Shakaa)

-7p^2 + 24p + 16 <0
(Is this right and if so where do i take it from here?)
-(7p^2 -24p -16) < 0

7p^2 - 24p -16 > 0 note reversal of sign

(7p+4)(p-4) > 0

When p > 4 both brackets are positive so satisified
When p < -4/7 both brackets are negative so satisfied

When p lies between these critical values, first bracket is +ve, second is -ve, so product is -ve.

### Related university courses

TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is supported by:
Updated: February 15, 2010
Today on TSR

### Top unis in Clearing

Tons of places at all these high-ranking unis

Poll
Useful resources

### Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

### How to use LaTex

Writing equations the easy way

### Study habits of A* students

Top tips from students who have already aced their exams