# solve the inequality

• Apr 30th 2009, 08:42 AM
solve 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

• Apr 30th 2009, 08:50 AM
e^(i*pi)
Quote:

solve the inequality

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

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

Either \$\displaystyle 3x^2-10x+3=0\$ or \$\displaystyle 2x^2-5x+2=0\$

\$\displaystyle 3x^2-10x+3=0\$ 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.

\$\displaystyle 2x^2-5x+2=0\$ 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
• Apr 30th 2009, 08:55 AM
VonNemo19
Hi
Looks good to me!(Clapping)
• Apr 30th 2009, 09:35 AM
Quote:

Originally Posted by e^(i*pi)
Either \$\displaystyle 3x^2-10x+3=0\$ or \$\displaystyle 2x^2-5x+2=0\$

\$\displaystyle 3x^2-10x+3=0\$ 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.

\$\displaystyle 2x^2-5x+2=0\$ 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

firstly, its not equal but < 0.

lets consider your first argument, acc to which;

Quote:

\$\displaystyle 3x^2-10x+3=0\$ 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.
but we need the solution to

\$\displaystyle 3x^2-10x+3<0\$

which i can rewrite as;

\$\displaystyle 3(x-1/3)(x-3)<0\$

which is true only when \$\displaystyle 1/3<x<3\$

& 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.
• Apr 30th 2009, 09:40 AM
now, lets just go with my textbook answer and put in a value of \$\displaystyle x<1/3\$ 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)
• Apr 30th 2009, 02:15 PM
stapel
Quote:

solve the inequality

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

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?

. . . . .\$\displaystyle 3x^2\, -\, 10x\, +\, 3\, <\, 0\$

. . . . .\$\displaystyle (3x\, -\, 1)(x\, -\, 3)\, <\, 0\$

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).

. . . . .\$\displaystyle 2x^2\, -\, 5x\, +\, 2\, >\, 0\$

. . . . .\$\displaystyle (2x\, -\, 1)(x\, -\, 2)\, >\, 0\$

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)
• Apr 30th 2009, 05:21 PM
Quote:

Originally Posted by stapel
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?

. . . . .\$\displaystyle 3x^2\, -\, 10x\, +\, 3\, <\, 0\$

. . . . .\$\displaystyle (3x\, -\, 1)(x\, -\, 3)\, <\, 0\$

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).

. . . . .\$\displaystyle 2x^2\, -\, 5x\, +\, 2\, >\, 0\$

. . . . .\$\displaystyle (2x\, -\, 1)(x\, -\, 2)\, >\, 0\$

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)

well all that we could just do with the wavy curve method which gives the answer which i got. so finally the textbook answer(x<1/3) is wrong i believe...
• Apr 30th 2009, 05:29 PM
mr fantastic
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