Math / Algebra

Question $[-/ 2$ Points] DETAILS MY NOTES SALGTRIG4 1.7.044.MI.
Solve the nonlinear inequality. Express the solution using interval notation. $x^{2}+2 x>3$

Studdy Solution

STEP 1

Assumptions
1. We need to solve the inequality $x^2 + 2x > 3$ .
2. We will express the solution in interval notation.
3. We will assume $x$ is a real number.

STEP 2

First, let's rewrite the inequality in standard form by moving all terms to one side of the inequality.
$x^2 + 2x - 3 > 0$

STEP 3

Next, we need to factor the quadratic expression $x^2 + 2x - 3$ . To do this, we look for two numbers that multiply to $-3$ and add to $2$ .

STEP 4

The factors of $-3$ that add up to $2$ are $3$ and $-1$ . Therefore, we can factor the quadratic expression as:
$(x + 3)(x - 1) > 0$

STEP 5

To solve the inequality $(x + 3)(x - 1) > 0$ , we need to determine the intervals where the product of the factors is positive. We do this by finding the critical points where each factor is zero.
$x + 3 = 0 \quad \Rightarrow \quad x = -3$ $x - 1 = 0 \quad \Rightarrow \quad x = 1$

STEP 6

The critical points divide the real number line into three intervals: $(-\infty, -3)$ , $(-3, 1)$ , and $(1, \infty)$ . We need to test a point in each interval to determine where the inequality holds.

STEP 7

Choose a test point in the interval $(-\infty, -3)$ , for example, $x = -4$ .
$(x + 3)(x - 1) = (-4 + 3)(-4 - 1) = (-1)(-5) = 5 > 0$
Since the product is positive, the inequality holds in the interval $(-\infty, -3)$ .

STEP 8

Choose a test point in the interval $(-3, 1)$ , for example, $x = 0$ .
$(x + 3)(x - 1) = (0 + 3)(0 - 1) = 3(-1) = -3 < 0$
Since the product is negative, the inequality does not hold in the interval $(-3, 1)$ .

STEP 9

Choose a test point in the interval $(1, \infty)$ , for example, $x = 2$ .
$(x + 3)(x - 1) = (2 + 3)(2 - 1) = 5(1) = 5 > 0$
Since the product is positive, the inequality holds in the interval $(1, \infty)$ .

STEP 10

Combine the intervals where the inequality holds. The solution to the inequality is the union of the intervals $(-\infty, -3)$ and $(1, \infty)$ .

STEP 11

Express the solution in interval notation.
$(-\infty, -3) \cup (1, \infty)$
The solution to the inequality $x^2 + 2x > 3$ in interval notation is $(-\infty, -3) \cup (1, \infty)$ .

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Question $[-/ 2$ Points] DETAILS MY NOTES SALGTRIG4 1.7.044.MI.
Solve the nonlinear inequality. Express the solution using interval notation. $x^{2}+2 x>3$

Studdy Solution

STEP 1

Assumptions
1. We need to solve the inequality $x^2 + 2x > 3$ .
2. We will express the solution in interval notation.
3. We will assume $x$ is a real number.

STEP 2

First, let's rewrite the inequality in standard form by moving all terms to one side of the inequality.
$x^2 + 2x - 3 > 0$

STEP 3

Next, we need to factor the quadratic expression $x^2 + 2x - 3$ . To do this, we look for two numbers that multiply to $-3$ and add to $2$ .

STEP 4

The factors of $-3$ that add up to $2$ are $3$ and $-1$ . Therefore, we can factor the quadratic expression as:
$(x + 3)(x - 1) > 0$

STEP 5

To solve the inequality $(x + 3)(x - 1) > 0$ , we need to determine the intervals where the product of the factors is positive. We do this by finding the critical points where each factor is zero.
$x + 3 = 0 \quad \Rightarrow \quad x = -3$ $x - 1 = 0 \quad \Rightarrow \quad x = 1$

STEP 6

The critical points divide the real number line into three intervals: $(-\infty, -3)$ , $(-3, 1)$ , and $(1, \infty)$ . We need to test a point in each interval to determine where the inequality holds.

STEP 7

Choose a test point in the interval $(-\infty, -3)$ , for example, $x = -4$ .
$(x + 3)(x - 1) = (-4 + 3)(-4 - 1) = (-1)(-5) = 5 > 0$
Since the product is positive, the inequality holds in the interval $(-\infty, -3)$ .

STEP 8

Choose a test point in the interval $(-3, 1)$ , for example, $x = 0$ .
$(x + 3)(x - 1) = (0 + 3)(0 - 1) = 3(-1) = -3 < 0$
Since the product is negative, the inequality does not hold in the interval $(-3, 1)$ .

STEP 9

Choose a test point in the interval $(1, \infty)$ , for example, $x = 2$ .
$(x + 3)(x - 1) = (2 + 3)(2 - 1) = 5(1) = 5 > 0$
Since the product is positive, the inequality holds in the interval $(1, \infty)$ .

STEP 10

Combine the intervals where the inequality holds. The solution to the inequality is the union of the intervals $(-\infty, -3)$ and $(1, \infty)$ .

STEP 11

Express the solution in interval notation.
$(-\infty, -3) \cup (1, \infty)$
The solution to the inequality $x^2 + 2x > 3$ in interval notation is $(-\infty, -3) \cup (1, \infty)$ .

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Question[−/2[-/ 2[−/2 Points] DETAILS MY NOTES SALGTRIG4 1.7.044.MI.Solve the nonlinear inequality. Express the solution using interval notation. x2+2x>3x^{2}+2 x>3x2+2x>3

Studdy Solution

STEP 1

Assumptions1. We need to solve the inequality x2+2x>3x^2 + 2x > 3x2+2x>3.2. We will express the solution in interval notation.3. We will assume xxx is a real number.

STEP 2

First, let's rewrite the inequality in standard form by moving all terms to one side of the inequality.x2+2x−3>0x^2 + 2x - 3 > 0x2+2x−3>0

STEP 3

Next, we need to factor the quadratic expression x2+2x−3x^2 + 2x - 3x2+2x−3. To do this, we look for two numbers that multiply to −3-3−3 and add to 222.

STEP 4

The factors of −3-3−3 that add up to 222 are 333 and −1-1−1. Therefore, we can factor the quadratic expression as:(x+3)(x−1)>0(x + 3)(x - 1) > 0(x+3)(x−1)>0

STEP 5

STEP 6

The critical points divide the real number line into three intervals: (−∞,−3)(-\infty, -3)(−∞,−3), (−3,1)(-3, 1)(−3,1), and (1,∞)(1, \infty)(1,∞). We need to test a point in each interval to determine where the inequality holds.

STEP 7

STEP 8

STEP 9

Choose a test point in the interval (1,∞)(1, \infty)(1,∞), for example, x=2x = 2x=2.(x+3)(x−1)=(2+3)(2−1)=5(1)=5>0(x + 3)(x - 1) = (2 + 3)(2 - 1) = 5(1) = 5 > 0(x+3)(x−1)=(2+3)(2−1)=5(1)=5>0Since the product is positive, the inequality holds in the interval (1,∞)(1, \infty)(1,∞).

STEP 10

Combine the intervals where the inequality holds. The solution to the inequality is the union of the intervals (−∞,−3)(-\infty, -3)(−∞,−3) and (1,∞)(1, \infty)(1,∞).

STEP 11

Express the solution in interval notation.(−∞,−3)∪(1,∞)(-\infty, -3) \cup (1, \infty)(−∞,−3)∪(1,∞)The solution to the inequality x2+2x>3x^2 + 2x > 3x2+2x>3 in interval notation is (−∞,−3)∪(1,∞)(-\infty, -3) \cup (1, \infty)(−∞,−3)∪(1,∞).

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Question $[-/ 2$ Points] DETAILS MY NOTES SALGTRIG4 1.7.044.MI.
Solve the nonlinear inequality. Express the solution using interval notation. $x^{2}+2 x>3$

Assumptions
1. We need to solve the inequality $x^2 + 2x > 3$ .
2. We will express the solution in interval notation.
3. We will assume $x$ is a real number.

First, let's rewrite the inequality in standard form by moving all terms to one side of the inequality.
$x^2 + 2x - 3 > 0$

Next, we need to factor the quadratic expression $x^2 + 2x - 3$ . To do this, we look for two numbers that multiply to $-3$ and add to $2$ .

The factors of $-3$ that add up to $2$ are $3$ and $-1$ . Therefore, we can factor the quadratic expression as:
$(x + 3)(x - 1) > 0$

The critical points divide the real number line into three intervals: $(-\infty, -3)$ , $(-3, 1)$ , and $(1, \infty)$ . We need to test a point in each interval to determine where the inequality holds.

Choose a test point in the interval $(1, \infty)$ , for example, $x = 2$ .
$(x + 3)(x - 1) = (2 + 3)(2 - 1) = 5(1) = 5 > 0$
Since the product is positive, the inequality holds in the interval $(1, \infty)$ .

Combine the intervals where the inequality holds. The solution to the inequality is the union of the intervals $(-\infty, -3)$ and $(1, \infty)$ .

Express the solution in interval notation.
$(-\infty, -3) \cup (1, \infty)$
The solution to the inequality $x^2 + 2x > 3$ in interval notation is $(-\infty, -3) \cup (1, \infty)$ .