To solve an inequality with modulus
-
A new topic is the solving of an inequality with modulus introduced in the new syllabus 2008 of Additional Mathematics. Hence, there are no past year exam questions on this topic in the ten years series.
In the previous syllabus, students are required to sketch the inequality and state the number of solutions but in the new syllabus, students are required to solve an inequality with modulus.
Question 1
Without using a graphics calculator, solve (not sketch) the inequality
l x^2 - 3x - 7 l < 3.
-
x^2 - 3x - 7 < 3
x^2 - 3x - 7 <-3
x^2 - 3x -10 <0
x^2 - 3x - 4 < 0
factorise get the answer and check for validity ~
(X-5)(X+2) <0
X = 5 or -2
(X-4)(X+1) <0
X= 4 Or -1
X<0 , Therefore answer is -1, -2 ...
nt sure ~ -
It would be stupid to listen to someone who barely passes A-Maths ~
-
|x^2 - 3x - 7| < 3
-3 < x^2 - 3x - 7 < 3
-
X²-3x-7 < 3 x²-3x-7<-3
X² - 3x – 10 < 0 x²- 3x – 4 <0
(x-5)(x+2) <0 (x-4) (x+1) <0
X = 5, -2, 4, -1
hope can bring u some where bahs...
-
Hi Yong93, Maurizio13 and fathernature,
When we solve an inequality with modulus, the answer will still be inequalities
ie x > a, x < b, x > - c, x < - d, b < x < a, - d < x < - c, -c < x < a etc
Thank you for your kind attention.
Regards,
ahm97sic
-
Yes. Thanks for your advice, I know.
-
l x^2 - 3x - 7 l < 3
Step 1: Remove the modulus sign by recognising that the function could take on values from -3 to 3 such that the modulus is less than 3.
-3 < x^2 - 3x - 7 < 3
Step 2: Solve the 2 sides - factorising, etc.
0 < x^2 - 3x - 4 and x^2 - 3x - 10 < 0
0 < (x - 4) (x +1) and (x - 5) (x + 2) < 0
Step 3: Determine the range of values of x
__\________/___
-1 \ / 4
\ /
\ ___/For this inequality to be more than zero, x < -1 or x > 4
__\________/___
-2 \ / 5
\ /
\ ___/For this inequality to be less than zero, -2 < x < 5
Combining these 2 inequalities, you get -2 < x < -1 or 4 < x < 5
I hope I'm correct.
-
solution set: -1 < x < -2 OR 4 < x < 5 -
....
-
I
Hope my solution helps. =)
-
Originally posted by onigiri:
I
Hope my solution helps. =)
Hi Onigirl,
Yes, it is the solution to the question. You have given a detailed and good explanation for the solution.
Thank you for your kind attention.
Regards,
ahm97sic
PS : Hi Ms Onigirl, can you kindly tell me the software which you use to type and put up your answer in the forum. Thanks !
-
hello there! haha, glad to know that i've been of some help. =)
and sorry, just realised that there's actually a typo-error in one of the answers. it should be -2 < x < -1, not -2 < x < 1.
anyways, i just used powerpoint to type out my solution and draw out those diagrams.
after that, i printscreen-ed the slide and saved it as a picture, then uploaded it onto this forum.
p/s: it is onigiri, not onigirl. hahaha!
-
Originally posted by onigiri:
hello there! haha, glad to know that i've been of some help. =)
and sorry, just realised that there's actually a typo-error in one of the answers. it should be -2 < x < -1, not -2 < x < 1.
anyways, i just used powerpoint to type out my solution and draw out those diagrams.
after that, i printscreen-ed the slide and saved it as a picture, then uploaded it onto this forum.
p/s: it is onigiri, not onigirl. hahaha!
Hi Onigiri,
Thank you very much.
Regards,
ahm97sic