Please Help Me.

Dear all, I have quite a few questions regarding A-maths that I want to ask. I am really confused on how to do these questions and I have tried hard to do them but the answers I got are wrong. Please Help Me. I sincerely want to do well and understand A-maths. Thanks.


1) Given that , for all values of x, find the value of each of A, B and C.
[Ans: A = 2, B = −3, C = 3]


2) For all values of x, . Find the value of A, B and C.
[Ans: A = 3, B = 2, C = 1]


3) Given that leaves a remainder of −4 when divided by x − a, find the possible values of a.
[Ans: A = −2, , or ]


4) Find the range of values of x for which x(2x + 5) > 12.
[Ans: x < −4 or x > ]


5) (a) Find the range of value of x for which x^2+7x-9 > 8x-3

(b) Find the range of value of c for which x^2+7x-9 > 8x+c , for all values of x.

[Ans: (a) −2 < x < 3 (b) c < -(37/4) ]


6) Find the value of x for which 2x^2 > 3x+14

[Ans: x < −2 or x > 3 (1/4) ]



I would really be glad if anyone of you help me with the above questions. Please Help. Thanks a lot. :)

Please help me. :cry:

Yunie,

Your questions appear to be incomplete.

For example,

Quote:

---

1) Given that , for all values of x, find the value of each of A, B and C.
[Ans: A = 2, B = −3, C = 3]

---

Given that what???

If you cut & pasted that text, perhaps some of the source wasn't text, but images? Perhaps you could fill in the missing information by hand.

Then it'd be much easier to help.

:wave:
-Lou

Sorry, but perhaps you could help me with the other questions first? Thanks. =)

For the part 4.

Simplify the expression as follows.

X(2X + 5) > 12

2X² + 5X - 12 > 0

2X² + 8X - 3X - 12 > 0

2X(X + 4) - 3(X + 4) > 0

(2X - 3)(X + 4) > 0

Assume,

(2X - 3) = 0 and (X + 4) = 0

So, X = 1.5 and X = -4

Now you have to put them in number line and check in which area it is satisfied. It's like this,

if X < -4 then (2X - 3)(X + 4) > 0
if X > -4 and X < 1.5 then (2X - 3)(X + 4) < 0
if X > 1.5 then (2X - 3)(X + 4) > 0

Therefore, the answer is (X < -4) or (X > 1.5)

Now I think you can do the rest. If you want I'll do it part by part later. Just try it.

Quote:

---

Originally Posted by eranga262154

For the part 4.

Simplify the expression as follows.

X(2X + 5) > 12

2X² + 5X - 12 > 0

2X² + 8X - 3X - 12 > 0

2X(X + 4) - 3(X + 4) > 0

(2X - 3)(X + 4) > 0

Assume,

(2X - 3) = 0 and (X + 4) = 0

So, X = 1.5 and X = -4

Now you have to put them in number line and check in which area it is satisfied. It's like this,

if X < -4 then (2X - 3)(X + 4) > 0
if X > -4 and X < 1.5 then (2X - 3)(X + 4) < 0
if X > 1.5 then (2X - 3)(X + 4) > 0
Therefore, the answer is (X < -4) or (X > 1.5)

Now I think you can do the rest. If you want I'll do it part by part later. Just try it.

---

How do you get the 8x? And, I don't understand about the highlighted parts. :confused: :ehh:

For part 5 a,

Simplify the expression first,

X² + 7X - 9 > 8X - 3

X² - X - 6 > 0

X² - 3X + 2X - 6 > 0

X(X - 3) + 2(X - 3) > 0

(X + 2)(X - 3) > 0

Assume,

(X + 2) = 0 and (X - 3) = 0
So, X = -2 and X = 3

Mark them on a number line, and find the proper range.

if X < -2 then (X + 2)(X - 3) > 0
if X > -2 and X < 3 then (X + 2)(X - 3) < 0
if X < 3 then (X + 2)(X - 3) > 0

So the answer is, (X < -2) or (X > 3).

Your give answer(-2 < X < 3) is completely wrong. Say according to your answer X = 0. Substitute it on the original expression.

X² + 7X - 9 > 8X - 3

X=0;
0² + 7(0) - 9 > 8(0) - 3
-9 > -3

It can't be never happened.



For part b,

X² + 7X - 9 > 8X + C

X² - X - (9 + C) > 0

To satisfied this condition, either X < -2 or X > 3

if X < -2,

(-2)² - (-2) - (9 + C) > 0
4 + 4 - 9 - C > 0
- 1 - C > 0
C < -1 -------------------- (A)


if X > 3

(3)² - (3) - (9 + C) > 0
9 - 3 - 9 - C > 0
- 3 - C > 0
C < -3 -------------------- (B)

From (A) and (B), C < -3 is the answer. Check with your book or any referring materials. I think I'm in correct way.

Ok I'll start with basis.

Say the quadric equation in the form of,

aX² + bX + C = 0

here, b should be written in summation of the product of a and c, with the sign. That is if b is a negative number either one of the summation number should be negative, not both. If b is a positive number the the both summation numbers should be positive or negative.

So in your question a = +2, b = +5, c = -12
ie: ab = -24 = +8 * -3

Hence,
( +5X ) = ( +8X ) * ( - 3X)

For the second part assume a value for X.

Code:

---

if X < -4 then (2X - 3)(X + 4) > 0
if X > -4 and X < 1.5 then (2X - 3)(X + 4) < 0
if X > 1.5 then (2X - 3)(X + 4) > 0

---

For line 1, say x = -5, then the expression is hold a positive value.

Same thing for other two lines.

Are you familiar with the number line. Behavior of the part 4 is as follows.

Lets get more freindly.Imagine the table below before and after filling.We know that our results from the Quadratic expression are x = 1.5 and x = 4
Hence to get the range, we should test for a positive result when x < -4, and when -4 < x < 1.5 and finally when x > 1.5. To appreciate this lets draw a table looking like the one below.

x < 4 -4 < x < 1.5 x > 1.5
(2x-3)
(x + 4)
(2x - 3)(x + 4)


Then after the test, we get.

x < 4 -4 < x < 1.5 x > 1.5
(2x-3) Positive Negative Positive
(x + 4) Positive Positive Positive
(2x - 3)(x + 4) Positive Negative Positive

Hence the table tells it all that x < 4 and x > 1.5.

Exactly that what I put there using "if" statements. I think it is not difficult to understand.