SOLUTION: Hi, I'm studying for a test and having trouble solving the below problem: 2x^2 + 3x < 9 I'm assuming/hoping this is a completing the square problem? x^2 + 3/2x

SOLUTION: Hi, I'm studying for a test and having trouble solving the below problem: 2x^2 + 3x < 9 I'm assuming/hoping this is a completing the square problem? x^2 + 3/2x < 9/2 x^

Question 360016: Hi, I'm studying for a test and having trouble solving the below problem:
2x^2 + 3x < 9
I'm assuming/hoping this is a completing the square problem?
x^2 + 3/2x < 9/2
x^2 + 3/2x + 9/4 < 27/4
This is where my head starts to spin lol. I'm supposed to find the square of the quadratic on the left...there has got to be an easier way to do this! If not, then I'm really not sure how I would go about factoring that equation, so any guidance you can provide is greatly appreciated!
Found 2 solutions by scott8148, tam144:
Answer by scott8148(6628) (Show Source): You can put this solution on YOUR website!
subtracting 9 ___ 2x^2 + 3x - 9 < 0

factoring ___ (2x - 3)(x + 3) < 0
Answer by tam144(6) (Show Source): You can put this solution on YOUR website!
Always solve a quadratic inequality in 4 steps.
Step 1. Bring it to standard form: f(x) = 2x^2 + 3x - 9 < 0.
Step 2. Solve f(x) = 0. You can use any method you prefer. I use the new Diagonal Sum method. Roots have opposite signs. There are 3 probable root-pairs:
(-1/2, 9/1),(-3/1, 3/2),(-3/2, 3/1). The diagonal sum of the second set is -3 = -b. The 2 real roots are -3 and 3/2.
Or, you can solve it by the factoring ac method (You Tube). Find 2 number that their product is ac = -18, and their sum is b = 3. Proceed: [(-1, 18)(1, -18)(-2, 9)(2, -9)(-3, 6), OK]. Replace in the equation f(x) = 0 the quantity 3x by two quantities -3x and 6x.
2x^2 + 3x - 9 = 2x^2 - 3x + 6x - 9 = 0.
= 2x(x + 3)- 3(x +3) = (x + 3)(2x - 3). Solve the 2 binomials:
x + 3 = 0 ---> x = -3
2x - 3 = 0 ---> x = 3/2
Step 3. Solve the inequality f(x) < 0. Use the number line and test point method. Plot the 2 real roots -3 and 3/2 on the number line. Use the origin O as test point. Substitute x = 0 into the inequality. You get -9 < 0. It is true, then the origin O is on the true segment (-3, 3/2).
Step 4. Express the answer (solution set) of the inequality in the form of an open interval (-3, 3/2). The 2 end points -3 and 3/2 are not included in the solution set.
If in the inequality, there is an additional (=) sign (lesser or equal to), then the solution set is a closed interval [-3, 3/2]. The 2 end points -3 and 3/2 are included in the solution set.
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