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Old 8th April 2012, 03:45 PM   #1 (permalink)
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Discriminants and roots problem

Hi, I have got some problems which I could not solve:

Q1a)
Find the range of values of k for which the quadratic equation 2kx^2 + (8 - 4k)x + k + 1 = 0 has real roots. State the largest integer k for which this equation has no real roots.
Q1b)
If the equations px^2 + qx + 2r = 0 and rx^2 + px - q + 1 =0 each have equal real roots, find a relation between p and q.

A1a) k < or equal 1 or k > or equal 8, k is not equal 0; 7
A1b) 2p^3 = q^2 - q^3

Q2)
The curve y = ab^x passes through the points (0,5) and (2/3,5/4). Find the positive value of a and of b.

A2) a = 5, b = 1/8

Q3)a)
Find the non-zero value of k for which equation 3x^2 + 2kx - k = 0 has
i) 2 equal roots
ii) 2 real roots, one of which is the reciprocal of the other.
iii) 2 real roots, one of which is twice the other.

A3a)
i) -3
ii) -3
iii) -27/8

Q3b)
Find the value of k for which the equation 4(x + 1)(x - 4) = k has roots which differ by 8

A3b)
39

Q4)
The equation 2x^2 + 7x + a = 0 has roots A and B and 4x^2 + bx + 16 =0 has roots A^2 and B^2. Calculate the possible values of a and b.

A4)
a = 4, b = -33, or a = -4, b = -65

Q5)
The roots of a quadratic equation 3x^2 - square-root24 X - 2 = 0 are m and n where m > n. Find the value of 4/m - 2/n in the form square root a - square root b

A5) square root 108 - square root 6

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Old 8th April 2012, 05:00 PM   #2 (permalink)
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Re: Discriminants and roots problem

Are you trying to get someone do your homework here?

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Old 8th April 2012, 06:12 PM   #3 (permalink)
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Re: Discriminants and roots problem

dude, these are very simple sec 3 A-maths questions. If you're a student please attempt to read through the examples in your textbook and then try out the questions. If you just want free answers you will never ace your tests.

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Old 8th April 2012, 06:59 PM   #4 (permalink)
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Re: Discriminants and roots problem

You sure you can't solve them? They're really simple ! :/

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Old 8th April 2012, 07:19 PM   #5 (permalink)
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Re: Discriminants and roots problem

TS has been exposed~ Do it yourself before asking for help.

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Old 8th April 2012, 11:24 PM   #6 (permalink)
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Re: Discriminants and roots problem

Originally Posted by whitecorp View Post
Are you trying to get someone do your homework here?
No, I am only asking for guides

Originally Posted by rayozen View Post
dude, these are very simple sec 3 A-maths questions. If you're a student please attempt to read through the examples in your textbook and then try out the questions. If you just want free answers you will never ace your tests.
Originally Posted by KeaneLJC View Post
You sure you can't solve them? They're really simple ! :/
If it is simple, I wouldn't have wasted my time to post these questions already, I started A-Maths in Sec 5, inside the textbook, the examples only show basic questions, that is why I post these questions

Originally Posted by themightyblue View Post
TS has been exposed~ Do it yourself before asking for help.
I tried for days, I can't get the answer, Do you mean I should also post the answer that I attempted?

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Old 9th April 2012, 02:04 AM   #7 (permalink)
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Re: Discriminants and roots problem

Posting a long list of questions and their answers is not the way to ask for guidance. If you post your workings and show us where you are stuck, others would be more obliged to help. Peace.

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Old 9th April 2012, 06:43 AM   #8 (permalink)
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Re: Discriminants and roots problem

know of the formula to solve quadratic equation? x = -b +/- sqrt(b^2-4ac)/2a. also, matching power coefficients? they will help in Q1.

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Old 9th April 2012, 08:42 AM   #9 (permalink)
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Re: Discriminants and roots problem

Originally Posted by whitecorp View Post
Posting a long list of questions and their answers is not the way to ask for guidance. If you post your workings and show us where you are stuck, others would be more obliged to help. Peace.
Alright, I apologize

For Q1, I will try since there is a hint now

Here are how I attempt the questions:
For Q2, I do not know how to start working

For Q3,
a)i)
3x^2 + 2kx - k =0
(2k)^2 - 4(3)(-k) = 0
4k^2 - 12(-k) = 0
4k^2 + 12k = 0
4k = 0 or k + 3 = 0
k = 0 (rejected) or k = -3

for ii)
4k(k + 3) > or equal to 0
4k > or equal to 0, k + 3 > or equal to 0
k> or equal to 1/-3 <- is that what they mean by reciprocal?

iii)
4k(k + 3) > or equal to 0
4k(k + 3)^2 > or equal to 0
4k > or equal to 0, (k + 3)^2 > or equal to 0
(k^2 + 6k + 9) > or equal to 0 <- from here if I factorise out, it is still the same so I do not know what to do

b)
4(x+1)(x-4) = k
(4x+4)(x-4) = k
4x^2 - 12x - 16 - k = 0
4x^2 - 12x - 16 - k = 0
(-12)^2 - 4(4)(-16-k) > or equal to 0 - 8
144 + 256 + 16k > or equal to 0 - 8
400 + 16k > or equal to 0 - 8
400 < or equal to -16k - 8
408 < or equal to -16k
-25.2 < or equal to K

For Q4
2x^2 + 7x + 9 = 0
roots are A + B
a = 2, b = 7, c = a
A+B = 7/2
AB = a/2

4x^2 + bx + 16 = 0
a = 4, b = b, c = 16
A+B = -b/4
AB = 4

A^2+B^2
= (A+B)^2 - 2AB
= (-b/4)^2 - 2(4)
= b^2/16 - 81

A^2B^2
= (AB)^2
= (4)^2
= 16 <- from this point, I am not sure what to do next

For Q5
3x^2 - squareroot24x - 2 = 0
a = 3, b = -squareroot24, c=-2
m+n = squareroot24/3
mn= -2/3

4/m - 2/n
= 4n - 2m/mn
= (4n)^2 + (2m)^2 - 2mn/mn
= 16n^2 + 4m^2 - 2mn/mn
= 4(4n^2 + m^2) - 2mn/mn <- from this point, I do not know how to factorise 4(4n^2 + m^2) into somthing like (m + n) so that I can sub m + n into it

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Old 9th April 2012, 09:45 AM   #10 (permalink)
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Re: Discriminants and roots problem

Q1a) Quadratic equations are in the form of ax²+bx+c=0.
When the question says that the equation has real roots, it means the discriminant b²-4ac is greater than or equal to zero.

1b.) When quadratic equations have equal real roots, the discriminant b²-4ac=0. From the two given equations you can derive 2 new simultaneous equations involving only p and q through the use of aforementioned statement regarding the discriminant.

2.) You are given the equation of a curve and two points on the curve. First, substitute x=0 into the equation and you find that a=5. Since anything to the power of 0 = 1. Subsequently substitute the values of x=2/3, y=5/4 and a=5.

3ai.) Good job.

3aii.) The reciprocal of x is 1/x.

3aii and 3aiii.) You should use (x-a)(x-1/a)=0 and (x-a)(x-2a)=0 respectively.

3b.) Use (x-a)(x-(a+8))=0.

4.) You have careless mistake. your (A+B) should equal -7/2.

5.) (a-b)²=a²-2ab+b²=a²+b²-2ab=[(a+b)²-2ab]-2ab=(a+b)²-4ab

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