Answer :
Let's solve the inequality [tex]\( x^2 + 12x + 35 > 0 \)[/tex].
### Step 1: Factor the quadratic expression
The expression [tex]\( x^2 + 12x + 35 \)[/tex] can be factored into two binomials:
[tex]\[
x^2 + 12x + 35 = (x + 5)(x + 7)
\][/tex]
### Step 2: Determine the critical points
The critical points are the values of [tex]\( x \)[/tex] that make the expression equal to zero. So, set each factor equal to zero:
- [tex]\( x + 5 = 0 \)[/tex] gives [tex]\( x = -5 \)[/tex]
- [tex]\( x + 7 = 0 \)[/tex] gives [tex]\( x = -7 \)[/tex]
These critical points divide the number line into intervals.
### Step 3: Test intervals
We need to test intervals around the critical points to see where the product is greater than zero [tex]\((> 0)\)[/tex].
1. Interval: [tex]\( (-\infty, -7) \)[/tex]
For this interval, pick a test point, like [tex]\( x = -8 \)[/tex]:
[tex]\((x + 5)(x + 7) = (-8 + 5)(-8 + 7) = (-3)(-1) = 3\)[/tex]
Since the product is positive, the inequality holds.
2. Interval: [tex]\( (-7, -5) \)[/tex]
For this interval, pick a test point, like [tex]\( x = -6 \)[/tex]:
[tex]\((x + 5)(x + 7) = (-6 + 5)(-6 + 7) = (-1)(1) = -1\)[/tex]
Since the product is negative, the inequality does not hold.
3. Interval: [tex]\( (-5, \infty) \)[/tex]
For this interval, pick a test point, like [tex]\( x = 0 \)[/tex]:
[tex]\((x + 5)(x + 7) = (0 + 5)(0 + 7) = 5 \times 7 = 35\)[/tex]
Since the product is positive, the inequality holds.
### Step 4: Combine solutions from intervals
The solutions to the quadratic inequality [tex]\( (x + 5)(x + 7) > 0 \)[/tex] are the intervals where the product is positive:
- [tex]\( x < -7 \)[/tex]
- [tex]\( x > -5 \)[/tex]
Now let's look at the additional piece: [tex]\( x < 7 \)[/tex] or [tex]\( x > 5 \)[/tex].
### Step 5: Combine all conditions
Combine the results:
- [tex]\( x < -7 \)[/tex] fulfills [tex]\( x < 7 \)[/tex] portion.
- [tex]\( x > 5 \)[/tex] fulfills its own portion.
Therefore, the final solution combining both parts is:
- [tex]\( x < -7 \)[/tex]
- [tex]\( x > 5 \)[/tex]
In conclusion, the solution to the inequality, considering all conditions provided, is:
[tex]\[
x < -7 \quad \text{or} \quad x > 5
\][/tex]
### Step 1: Factor the quadratic expression
The expression [tex]\( x^2 + 12x + 35 \)[/tex] can be factored into two binomials:
[tex]\[
x^2 + 12x + 35 = (x + 5)(x + 7)
\][/tex]
### Step 2: Determine the critical points
The critical points are the values of [tex]\( x \)[/tex] that make the expression equal to zero. So, set each factor equal to zero:
- [tex]\( x + 5 = 0 \)[/tex] gives [tex]\( x = -5 \)[/tex]
- [tex]\( x + 7 = 0 \)[/tex] gives [tex]\( x = -7 \)[/tex]
These critical points divide the number line into intervals.
### Step 3: Test intervals
We need to test intervals around the critical points to see where the product is greater than zero [tex]\((> 0)\)[/tex].
1. Interval: [tex]\( (-\infty, -7) \)[/tex]
For this interval, pick a test point, like [tex]\( x = -8 \)[/tex]:
[tex]\((x + 5)(x + 7) = (-8 + 5)(-8 + 7) = (-3)(-1) = 3\)[/tex]
Since the product is positive, the inequality holds.
2. Interval: [tex]\( (-7, -5) \)[/tex]
For this interval, pick a test point, like [tex]\( x = -6 \)[/tex]:
[tex]\((x + 5)(x + 7) = (-6 + 5)(-6 + 7) = (-1)(1) = -1\)[/tex]
Since the product is negative, the inequality does not hold.
3. Interval: [tex]\( (-5, \infty) \)[/tex]
For this interval, pick a test point, like [tex]\( x = 0 \)[/tex]:
[tex]\((x + 5)(x + 7) = (0 + 5)(0 + 7) = 5 \times 7 = 35\)[/tex]
Since the product is positive, the inequality holds.
### Step 4: Combine solutions from intervals
The solutions to the quadratic inequality [tex]\( (x + 5)(x + 7) > 0 \)[/tex] are the intervals where the product is positive:
- [tex]\( x < -7 \)[/tex]
- [tex]\( x > -5 \)[/tex]
Now let's look at the additional piece: [tex]\( x < 7 \)[/tex] or [tex]\( x > 5 \)[/tex].
### Step 5: Combine all conditions
Combine the results:
- [tex]\( x < -7 \)[/tex] fulfills [tex]\( x < 7 \)[/tex] portion.
- [tex]\( x > 5 \)[/tex] fulfills its own portion.
Therefore, the final solution combining both parts is:
- [tex]\( x < -7 \)[/tex]
- [tex]\( x > 5 \)[/tex]
In conclusion, the solution to the inequality, considering all conditions provided, is:
[tex]\[
x < -7 \quad \text{or} \quad x > 5
\][/tex]
