Solve the Inequality question

Shu

Shu

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Hello, While on my test from the end of the semester this is another question I could not answer at all. Could somebody give me any advice? I'll add my work and the question to make it easier. Any help is appreciated, Thank you and have a nice night!

Question in exact words
"Solve The Inequality [math]2x^2 +5x-12>0[/math]"
 

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2x2 + 5x - 12 > 0 factors into (2x - 3)(x + 4) > 0. Solve each factor, test a point in each interval, and write your solution.
 
R.M. said:
2x2 + 5x - 12 > 0 factors into (2x - 3)(x + 4) > 0. Solve each factor, test a point in each interval, and write your solution.
Click to expand...
There are several ways You can complete this problem. If I were to finish this assignment, I would note that, when product of two numbers are positive, then
  • either both the factors are positive

  • or both the factors are negative, and

  • neither of the factors are equal to zero
From the information above, I would calculate the domain of the answer.
 
I would sketch a quick graph, labelling the x intercepts and read tne solution off the graph (ie what part of the graph lies above the x-axis).
 
A positive quadratic is positive to the left of the smaller x-intercept and to the right of the lager x-intercept. In between the two x-intercept the quadratic will be negative.

When will the quadratic equal to 0? What happens if the quadratic is negative?
 
Sinc that is a quadratic with positive leading coefficient, its graph is a parabola opening upward. If its vertex has negative y value then its value is negative for x between the zeros and postive for x outside the zeros.


R.M. has already told you that this factors into (2x- 3)(x- 4) so its zeros are x= 3/2 and x= -4. The inequality is satisfied for x< -4 and x> 3/2.

Or you could do what R.M. suggested, evaluate at a single point in each interval. The numbers x= -4 and x= 3/2 divide the number line into three intervals- x< -4, -4< x< 3/2, and x> 3/2. -5 is less than -4 and if x= -5 we have 2(-5)^2+ 5(-5)- 12= 50- 25- 12= 25- 12= 13> 0. 0 is between -4 and 3/2 and if x= 0 we have 2(0)^2+ 5(0)- 12= -12< 0. Finally, 2 is greater than 3/2 and if x= 2 we have 2(2)^3+ 5(2)- 12= 8+ 10- 12= 6> 0.

Yet another method- The product of two numbers is negative if and only if they have different sign. If they are either both positive or both negative then the product is positive.
2x- 3> 0 for x> 3/2 and x+ 4> 0 for x> -4. Those are both true if x> 3/2.
2x- 3< 0 for x< 3/2 and x+ 4< 0 for x< -4. Those are both true if x< -4.
Again, we have that 2x^2+ 5x- 12> 0 if x< -4 or if x> 3/2.
 
I still think, if the graph is easy to draw as it is in this case, then you can read the solution off the graph.
x-ints are -4 and 3/2. The parabola is shaped U. The graph lies above the x-axis (ie y>0) to the left of -4 (ie x<-4) and to the right of 3/2 (ie x>3/2).
Done!
 
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