solve the inequality

(3x^2-10x+3)(2x^2-5x+2)<0

i got 1/3<x<1/2 or 2<x<3

& the answer is x<1/3

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- April 30th 2009, 09:42 AMadhyetasolve the inequality
solve the inequality

(3x^2-10x+3)(2x^2-5x+2)<0

i got 1/3<x<1/2 or 2<x<3

& the answer is x<1/3 - April 30th 2009, 09:50 AMe^(i*pi)
Either or

has roots of 3 and 1/3 so x<3 and x<1/3. As 1/3 is less than 3 we can disregard the first solution.

has roots of 2 and 1/2. Since neither of those are less than 1/3 they can both be disregarded.

Therefore the solution is x<1/3 since all the others do not satisfy this one - April 30th 2009, 09:55 AMVonNemo19Hi
Looks good to me!(Clapping)

- April 30th 2009, 10:35 AMadhyeta
firstly, its not equal but < 0.

lets consider your first argument, acc to which;

Quote:

has roots of 3 and 1/3 so x<3 and x<1/3. As 1/3 is less than 3 we can disregard the first solution.

which i can rewrite as;

which is true only when

& acc. to you its true for x<1/3

same thing goes for the second.

Lets consider that you are right with that solution,

then any value of x<1/3 would satisfy 3(x-1/3)(x-3)<0

take any value, it wont, however a value BETWEEN 1/3 & 3 will. - April 30th 2009, 10:40 AMadhyeta
now, lets just go with my textbook answer and put in a value of for instance lets put x=-1

the expression comes out to be positive & hence, x=-1 isnt a solution to the inequality.

what do you ppl think?

(Wondering) - April 30th 2009, 03:15 PMstapel
For the product to be negative, one

*but not both*of the factors must be negative.

Where is 3x^2 - 10x + 3 less than zero? Assuming 3x^2 - 10x + 3 < 0, where is 2x^2 - 5x + 2 greater than zero? What is the overlap of these solutions?

. . . . .

. . . . .

From the equality, (3x - 1)(x - 3) = 0, we know that the above crosses the x-axis at x = 1/3 and at x = 3. From the graph of y = 3x^2 - 10x + 3, we know that the graph is below the x-axis (and thus that the inequality is negative) between these zeroes; that is, on the interval (1/3, 3).

. . . . .

. . . . .

From the equation, (2x - 1)(x - 2) = 0, we know the graph crosses the x-axis at x = 1/2 and at x = 2. From the graph of y = 2x^2 - 5x + 2, we know that the graph is above the x-axis (and thus that the inequality is positive) "on the ends"; that is, on the intervals (-infinity, 1/2) and (2, +infinity).

For both inequalities to have the signs we want, we need the intersection of these two solutions: [(-infinity, 1/2) union (2, +infinity)] intersect (1/3, 3).

What then is the solution-interval for this case?

Now what if you start with the inequalities on the factors going in the other direction?

(Wink) - April 30th 2009, 06:21 PMadhyeta
- April 30th 2009, 06:29 PMmr fantastic