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)
I don't understand the notation in your answer, this is how I would do this question. I will do the first one.
x^2 - 6x + 5 > 0
(x-1)(x-5) > 0
giving me "separators" at x=1 and x=5
So we have 3 regions on the number line
a)all values x<1
b) all values between 1 and 5
c) all values x>5
I then take an arbitrary value in each of these regions and test if the result is positive, not really caring about the actual answer but only its sign.
a) eg. let x=0 we get (-)(-) which is positive, so x<1 works
b) let x = 4, we get (+)(-) which is negative, so that region does not work
c) x=10 we get (+)(+) which is positive so x>5 works
therefore x<1 OR x>5 is the solution
To check your answers, we can solve each polynomial inequality step by step and graph the solution set on a number line.
1. x^2 - 6x + 5 > 0:
Factor the quadratic expression: (x - 1)(x - 5) > 0
Now, we need to determine the sign of this expression for different intervals of x. To do this, we can use test values.
Choose a value less than 1, like x = 0. Substitute it into the inequality: (0 - 1)(0 - 5) > 0
This gives us (-1)(-5) > 0, which is true.
Next, choose a value between 1 and 5, like x = 3: (3 - 1)(3 - 5) > 0
Simplifying, we get (2)(-2) > 0, which is false.
Lastly, choose a value greater than 5, like x = 6: (6 - 1)(6 - 5) > 0
By calculating, we find (5)(1) > 0, which is true.
From the test values, we can determine the solution set:
(-∞, 1) ∪ (5, ∞)
2. x^2 - 3x - 18 < 0:
Factor the quadratic expression: (x - 6)(x + 3) < 0
Now, test different intervals using test values:
Choose a value less than -3, like x = -4: (-4 - 6)(-4 + 3) < 0
Simplifying, we get (-10)(-1) < 0, which is true.
Next, choose a value between -3 and 6, like x = 0: (0 - 6)(0 + 3) < 0
Calculating this, we find (-6)(3) < 0, which is false.
Lastly, choose a value greater than 6, like x = 7: (7 - 6)(7 + 3) < 0
By calculation, we obtain (1)(10) < 0, which is false.
From the test values, we can determine the solution set:
(-∞, -3) U (6, ∞)
3. x^2 - 3x - 28 ≤ 0:
Factor the quadratic expression: (x - 7)(x + 4) ≤ 0
Again, test different intervals using test values:
Choose a value less than -4, like x = -5: (-5 - 7)(-5 + 4) ≤ 0
Simplifying, we get (-12)(-1) ≤ 0, which is false.
Next, choose a value between -4 and 7, like x = 0: (0 - 7)(0 + 4) ≤ 0
By calculating this, we find (-7)(4) ≤ 0, which is true.
Lastly, choose a value greater than 7, like x = 8: (8 - 7)(8 + 4) ≤ 0
By calculation, we obtain (1)(12) ≤ 0, which is false.
From the test values, we can determine the solution set:
(-∞, -4] U [7, ∞)
Therefore, your answers are correct.