Math  /  Algebra

QuestionSolve the following quadratic inequality: 3x2+5x2>03 x^{2}+5 x-2>0 Write your answer in interval notation. Note: Use oo for \infty and UU for union

Studdy Solution

STEP 1

What is this asking? We need to find the range of xx values that make the expression 3x2+5x23x^2 + 5x - 2 greater than zero! Watch out! Don't forget to consider the **sign** of the quadratic in each interval!

STEP 2

1. Factor the quadratic
2. Find the roots
3. Determine the sign in each interval

STEP 3

Let's **factor** our quadratic expression 3x2+5x23x^2 + 5x - 2.
We're looking for two numbers that multiply to (3)(2)=6(3) \cdot (-2) = -6 and add up to **5**.
Those lucky numbers are **6** and **-1**!

STEP 4

Rewrite the quadratic as 3x2+6xx23x^2 + 6x - x - 2.
Now, we can **factor by grouping**: 3x(x+2)1(x+2)3x(x+2) - 1(x+2).
This gives us (3x1)(x+2)(3x-1)(x+2).
Awesome!

STEP 5

To find the roots, we set our factored quadratic equal to zero: (3x1)(x+2)=0(3x-1)(x+2) = 0.
This gives us two **critical points**: 3x1=03x - 1 = 0 which means x=13x = \frac{1}{3}, and x+2=0x + 2 = 0 which means x=2x = -2.
These roots are where our quadratic might change its sign!

STEP 6

Now, let's **test** the sign of our quadratic in the three intervals created by our roots: x<2x < -2, 2<x<13-2 < x < \frac{1}{3}, and x>13x > \frac{1}{3}.

STEP 7

For x<2x < -2, let's pick x=3x = -3.
Plugging this into our factored form (3x1)(x+2)(3x-1)(x+2) gives us (3(3)1)(3+2)=(10)(1)=10(3(-3)-1)(-3+2) = (-10)(-1) = 10, which is **positive**!

STEP 8

For 2<x<13-2 < x < \frac{1}{3}, let's pick x=0x = 0.
Plugging this in gives us (3(0)1)(0+2)=(1)(2)=2(3(0)-1)(0+2) = (-1)(2) = -2, which is **negative**!

STEP 9

For x>13x > \frac{1}{3}, let's pick x=1x = 1.
Plugging this in gives us (3(1)1)(1+2)=(2)(3)=6(3(1)-1)(1+2) = (2)(3) = 6, which is **positive**!

STEP 10

We want the intervals where the quadratic is **greater than zero**, so that's x<2x < -2 and x>13x > \frac{1}{3}.

STEP 11

Our solution in interval notation is (,2)(13,)(-\infty, -2) \cup (\frac{1}{3}, \infty).
Boom!

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