"Find the set of values of x for which: 2x^2 − 11x + 12 < 0"?

Question

Could somebody please help me on maths questions like these? How do you know which way round the inequality signs go in your answer?

Answer 1

2x^2 - 11x + 12 < 0

To solve this, you must first obtain your critical values. Factor as normal to obtain

(2x - 3)(x - 4) < 0

Now, we solve the corresponding equation (2x - 3)(x - 4) = 0 to obtain our critical values. This implies (2x - 3) = 0 or (x - 4) = 0, corresponding to the solutions x = 3/2 or x = 4

Now that we have our critical values, we want to test the region *around* them for positivity/negativity. In particular, we want to test 3 regions:

a) x < 3/2 {i.e. the region less than 3/2}
b) 3/2 < x < 4 {the region between 3/2 and 4}
c) x > 4 {the region greater than 4}

All it takes is for us to test a *single* value within those regions, for the inequality (2x - 3)(x - 4) < 0. What we want are negative results, given that "< 0" means negative.

For the region (a), test x = 0. Then
[2(0) - 3] [0 - 4] = (-3)(-4) = 12, which is positive. Reject the solution set of x < 3/2.

For region (b), test x = 2. Then [2(2) - 3] [2 - 4] = (1)(-2) = -2, which is negative. Keep the solution set 3/2 < x < 4.

For region (c), test x = 100. We're going to get a positive solution, since, for (2x - 3) (x - 4), we're going to get (positive) times (positive), which will be positive. Reject the solution set x > 4.

Therefore, our answer is "the set of all x such that x is between 3/2 and 4", or, written formally,
{x | 3/2 < x < 4}

In interval notation, this is

x E (3/2, 4)

Where "E" means "is an element of."

Answer 2

2x^2 -11x+12<0
or,2x^2-8x-3x+12<0
or,2x(x-4)-3(x-4)<0
or,(2x-3)(x-4)<0
Since only a positive and a negative term on multiplication gives a negative term,so either,
1)(2x-3)>0 and (x-4)<0
or,2)(2x-3)<0 and (x-4)>0
for the 1st case,
2x-3>0
or,2x>3
or,x>(3/2) and
x-4<0
or,x<4
So,(3/2) For the 2nd case,
2x-3<0
or,2x<3
or,x<(3/2) and
x-4>0
or,x>4.But x cannot be less than (3/2) and greater than 4 at the same time.So,2nd case is not valid.
Hence,the set of values is (3/2)
always remember,
1)a +ve termXa negative term=a -ve term
2)a +ve termXa +ve term=a +ve term
3)a -ve termXa -ve term=a positive term

Answer 3

(2x-3)(x-4)<0
2x-3<0
2x<3
x<3/2

x-4<0
x<4

the inequality signs in your answer stay the same as the sign in the question