Math  /  Algebra

QuestionSolve the inequality by the test-point method. Write the solution in interval notation. x2+4x+3>0x^{2}+4 x+3>0 (3,1)(-3,-1) None of these answers (1,)(-1, \infty) (,3)(1,)(-\infty,-3) \cup(-1, \infty) (,3)(-\infty,-3)

Studdy Solution

STEP 1

1. The inequality x2+4x+3>0x^2 + 4x + 3 > 0 is a quadratic inequality.
2. The test-point method involves finding the roots of the corresponding equation and testing intervals around these roots.
3. The solution will be expressed in interval notation.

STEP 2

1. Find the roots of the quadratic equation.
2. Determine the intervals defined by these roots.
3. Use test points to determine which intervals satisfy the inequality.
4. Write the solution in interval notation.

STEP 3

Find the roots of the quadratic equation x2+4x+3=0x^2 + 4x + 3 = 0. To do this, factor the quadratic expression:
x2+4x+3=(x+1)(x+3)=0 x^2 + 4x + 3 = (x + 1)(x + 3) = 0

STEP 4

Set each factor equal to zero to find the roots:
x+1=0x=1 x + 1 = 0 \quad \Rightarrow \quad x = -1 x+3=0x=3 x + 3 = 0 \quad \Rightarrow \quad x = -3

STEP 5

Determine the intervals defined by the roots x=3x = -3 and x=1x = -1. These roots divide the number line into three intervals:
1. (,3)(- \infty, -3)
2. (3,1)(-3, -1)
3. (1,)(-1, \infty)

STEP 6

Use test points from each interval to determine which intervals satisfy the inequality x2+4x+3>0x^2 + 4x + 3 > 0.
- For the interval (,3)(- \infty, -3), choose a test point, e.g., x=4x = -4: $ (-4)^2 + 4(-4) + 3 = 16 - 16 + 3 = 3 > 0 \] This interval satisfies the inequality.
- For the interval (3,1)(-3, -1), choose a test point, e.g., x=2x = -2: $ (-2)^2 + 4(-2) + 3 = 4 - 8 + 3 = -1 < 0 \] This interval does not satisfy the inequality.
- For the interval (1,)(-1, \infty), choose a test point, e.g., x=0x = 0: $ (0)^2 + 4(0) + 3 = 3 > 0 \] This interval satisfies the inequality.

STEP 7

Write the solution in interval notation. The intervals that satisfy the inequality are (,3)(- \infty, -3) and (1,)(-1, \infty).
The solution is:
(,3)(1,) (-\infty, -3) \cup (-1, \infty)
The correct answer is:
(,3)(1,) (-\infty, -3) \cup (-1, \infty)

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