please check my answers. thanxs!
Solve the polynomial inequality and graph the solution set on a number line. Express the solution set in interval notation.
x^2 - 6x + 5 > 0
-I got (-ininity,1) u (5, infinity)
x^2 - 3x - 18 < 0
-I got (-infinity,-3) u (6, infinity)
x^2 - 3x - 28 ¡Ü 0
-I got (-ininity,-4] u [7,infinity)
To check the answers, we need to solve the given polynomial inequalities and graph the solution sets on the number line:
1. x^2 - 6x + 5 > 0
To solve this inequality, we can factor the quadratic equation:
(x - 1)(x - 5) > 0
The critical points are where the expression equals 0: x - 1 = 0 OR x - 5 = 0
Solving for x, we get x = 1 or x = 5.
Now we can check the sign of the expression for three intervals:
For x < 1: If we select x = 0, the expression is positive: (0 - 1)(0 - 5) = 5 > 0
For 1 < x < 5: If we select x = 2, the expression is negative: (2 - 1)(2 - 5) = -1 > 0
For x > 5: If we select x = 6, the expression is positive: (6 - 1)(6 - 5) = 5 > 0
From this analysis, we can see that the solutions are x < 1 and x > 5. Therefore, the solution set in interval notation is (-∞, 1) ∪ (5, ∞).
2. x^2 - 3x - 18 < 0
To solve this inequality, we can factor the quadratic equation:
(x - 6)(x + 3) < 0
The critical points are where the expression equals 0: x - 6 = 0 OR x + 3 = 0
Solving for x, we get x = 6 or x = -3.
Now we can check the sign of the expression for three intervals:
For x < -3: If we select x = -4, the expression is positive: (-4 - 6)(-4 + 3) = 10 > 0
For -3 < x < 6: If we select x = 0, the expression is negative: (0 - 6)(0 + 3) = -18 < 0
For x > 6: If we select x = 7, the expression is positive: (7 - 6)(7 + 3) = 10 > 0
From this analysis, we can see that the solution is -3 < x < 6. Therefore, the solution set in interval notation is (-3, 6).
3. x^2 - 3x - 28 ≤ 0
To solve this inequality, we can factor the quadratic equation:
(x - 7)(x + 4) ≤ 0
The critical points are where the expression equals 0: x - 7 = 0 OR x + 4 = 0
Solving for x, we get x = 7 or x = -4.
Now we can check the sign of the expression for three intervals:
For x < -4: If we select x = -5, the expression is positive: (-5 - 7)(-5 + 4) = 12 > 0
For -4 < x < 7: If we select x = 0, the expression is negative: (0 - 7)(0 + 4) = -28 < 0
For x > 7: If we select x = 8, the expression is positive: (8 - 7)(8 + 4) = 36 > 0
From this analysis, we can see that the solution is -∞ < x ≤ -4 or 7 ≤ x < ∞. Therefore, the solution set in interval notation is (-∞, -4] ∪ [7, ∞).
It seems like your answers are correct for all three polynomial inequalities.
To check your answers, we need to solve each polynomial inequality and graph the solution set on a number line.
1. x^2 - 6x + 5 > 0:
Start by factoring the quadratic expression or solving it by using the quadratic formula. In this case, the factors of the quadratic expression are (x - 1)(x - 5). Set each factor greater than zero to find the intervals where the expression is positive.
(x - 1) > 0: Adding 1 to both sides, x > 1.
(x - 5) > 0: Adding 5 to both sides, x > 5.
Now, graph the solution set on a number line. Place open circles at 1 and 5 since the inequality is strictly greater than.
-----o------o------
Since the inequality is > 0, shade the regions to the right of the open circles:
<=======================>
Expressing the solution set in interval notation:
(1, ∞)
2. x^2 - 3x - 18 < 0:
Factor the quadratic expression: (x - 6)(x + 3) < 0.
Now, set each factor less than zero to find the intervals where the expression is negative.
(x - 6) < 0: x < 6.
(x + 3) < 0: x < -3.
Graph the solution set on a number line with open circles at -3 and 6:
-----o------o------
Since the inequality is < 0, shade the region between -3 and 6.
<----o=======o----->
Expressing the solution set in interval notation:
(-3, 6)
3. x^2 - 3x - 28 ≤ 0:
Factor the quadratic expression: (x + 4)(x - 7) ≤ 0.
Now, set each factor less than or equal to zero to find the intervals where the expression is non-positive.
(x + 4) ≤ 0: x ≤ -4.
(x - 7) ≤ 0: x ≤ 7.
Graph the solution set on a number line with closed circles at -4 and 7:
----o=======o-------
Since the inequality is ≤ 0, shade the region between -4 and 7.
<----o=======o----->
Expressing the solution set in interval notation:
[-4, 7]