Maths - Equations and Inequalities

bonkysleuth SG

Hi, I have difficulties in 3 questions.

1. Solve the inequality |x-2| < 3x

I did obtain the answer but I have to reject one of them for reason(s) unknown.

I used the formula for modulus inequality where if |x|<a where a is more than 0, -a <x<a

So in this case I have -3x < |x-2| <3x

I have answers x>0.5 and x > -1 (rej). Why?

2. Solve the inequality 2x-4 / |x-1| < 1

I multiply the modulus term over to to '1' on the RHS since modulus function is always +ve and got 2x-4 < |x-1|

|x-1| > 2x-4

Again, I used the rule where if |x| > a, then x< -a or x > a

Now I have x-1 < -2x + 4 or x-1 > 2x - 4

Solving them, I have the answers x< 5/3 or x<3, which are very different from the exact answers given.

And I noticed something, if the 'x' term is on both the LHS and RHS of the equality, am I NOT allowed to use the modulus rule for inequality? It always gets me wrong answers or answers which I do not reject for no reason, just as the 2 questions above have proven.

3. Find the range(s) of x for which |2x-3| <x^2

So there is

2x - 3 < x^2

x^2-2x+3 >0 (rearrange) OR -2x + 3 < x^2

For this part,

D = (-2)^2 - 4(3) = -8

Since coeff. of x^2 is +ve and D<0,

y= x^2 - 2x +3 is always above x axis

and x^2 - 2x +3 > 0

But we can obtain no solution from this equation even though both the end result proves that the equation is > 0. So as long as D<0, I need not find solution from that particular equation, do I? I can get answers from the other equation for this question, just want to clarify this equation.

wee_ws SG

Hi,

For Q1, x > 0.5 and x > -1 imply x > 0.5 as a result of intersection of the two.

For Q2, x < 3 or x < 5/3 imply x < 3, as a result of union of the two.

For Q3, the graphical approach is more appropriate as it allows for better visualisation.

Solving Q3 via the analytical method is also possible, although one must be careful:

|2x - 3| < x^2

=> -x^2 < 2x - 3 < x^2

=> -x^2 < 2x - 3 and 2x - 3 < x^2

=> x^2 + 2x - 3 > 0 and x^2 - 2x + 3 > 0

=> x < -3 or x > 1 and (x - 1)^2 + 2 > 0, by completing the square.

=> x < -3 or x > 1 and x is any real number

=> x < -3 or x > 1, as a result of intersection of the two.

Thanks!

P.S. The solution to an inequality may be the entire real set (as we have just seen) or the empty set at times, e.g. no real values of x will satisfy the inequality

x^2 - 2x + 3 < 0.

Watch out for these two special cases, which Cambridge has asked in the past.

Cheers,
Wen Shih

bonkysleuth SG

Hi,

For Q1, x > 0.5 and x > -1 imply x > 0.5 as a result of intersection of the two.

For Q2, x < 3 or x < 5/3 imply x < 3, as a result of union of the two.

For Q3, the graphical approach is more appropriate as it allows for better visualisation.

Solving Q3 via the analytical method is also possible, although one must be careful:

|2x - 3| < x^2

=> -x^2 < 2x - 3 < x^2

=> -x^2 < 2x - 3 and 2x - 3 < x^2

=> x^2 + 2x - 3 > 0 and x^2 - 2x + 3 > 0

=> x < -3 or x > 1 and (x - 1)^2 + 2 > 0, by completing the square.

=> x < -3 or x > 1 and x is any real number

=> x < -3 or x > 1, as a result of intersection of the two.

Thanks!

P.S. The solution to an inequality may be the entire real set (as we have just seen) or the empty set at times, e.g. no real values of x will satisfy the inequality

x^2 - 2x + 3 < 0.

Watch out for these two special cases, which Cambridge has asked in the past.

Cheers,
Wen Shih

WHy is it that for Qn 2 we consider the union instead of intersection? It is a OR not AND. And for question 2, there's another set of answer which I couldnt obtain through the method i have used... i can't remember the answer, but i';; find out and repost again

wee_ws SG

Hi,

It is important and necessary to remember these rules:

- AND: consider intersection;

- OR: consider union.

{x is real such that x < 3} is a bigger set than {x is real such that x < 5/3}, so the union of both gives us the bigger set. Draw the number line and include both sets and you'll be able to understand better.

Alternatively, draw the graphs of y = (2x - 4) / |x-1| (with GC, of course, since it is a non-standard function) and y = 1 and see what happens.

A very practical tip: When the analytical method is not clear to you (because you can't see), use the graphical method to check (because it allows one to visualise).

Thanks.

Cheers,
Wen Shih