SOLUTION: Solve the following inequalities :
a) x^2 - 5x + 4 > 0
b) k(2k-3) < k^2 - 2
c) x^2 - 5x - 14 > 0
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Question 1200141: Solve the following inequalities :
a) x^2 - 5x + 4 > 0
b) k(2k-3) < k^2 - 2
c) x^2 - 5x - 14 > 0
Found 3 solutions by math_tutor2020, greenestamps, josgarithmetic:
Answer by math_tutor2020(3817) (Show Source): You can put this solution on YOUR website!
I'll do the first problem to get you started.
Let's solve the associated equation
x^2 - 5x + 4 = 0
So,
x^2 - 5x + 4 = 0
(x-1)(x-4) = 0
x-1 = 0 or x-4 = 0
x = 1 or x = 4
The quadratic formula is an alternative pathway.
The roots of x^2 - 5x + 4 are x = 1 and x = 4
Draw out a number line to plot 1 and 4 on it.
Label the following regions
A: stuff to the left of 1
B: stuff between 1 and 4
C: stuff to the right of 4
Pick a value from region A.
I'll select x = 0
Plug it into x^2 - 5x + 4 and simplify to get...
x^2 - 5x + 4
0^2 - 5*0 + 4
4
We arrive at a positive value.
Therefore, x = 0 makes x^2 - 5x + 4 > 0 to be true
Any other value in region A will arrive at this same conclusion.
Region A is part of the solution set.
Now onto region B.
Let's pick x = 2 as a representative value from this region 1 < x < 4
x^2 - 5x + 4
2^2 - 5*2 + 4
4 - 10 + 4
-2
The result is negative, which means x^2 - 5x + 4 > 0 would be false for any x value in region B.
Region B is NOT part of the solution set.
Lastly region C, which is x > 4
Let's pick x = 5
x^2 - 5x + 4
5^2 - 5*5 + 4
25 - 25 + 4
4
Like with region A, we get a positive result to show x^2 - 5x + 4 > 0 would be true for x values in this interval.
Region C is part of the solution set.
We found that
Region A, x < 1, is part of the solution set
Region C, x > 4, is also part of the solution set
Collectively the entire solution set is x < 1 or x > 4
We can write this in interval notation to get (-infinity, 1) U (4 < infinity)
The U symbol is the union operator. It glues together the two intervals in an "or" fashion.
We are either in the interval (-infinity, 1) OR we are in the interval (4 < infinity). We cannot be in both intervals at the same time.
Answer as inequalities: x < 1 or x > 4
Answer as interval notation: (-infinity, 1) U (4 < infinity)
Similar question is found here
https://www.algebra.com/algebra/homework/Inequalities/Inequalities.faq.question.1200712.html
Answer by greenestamps(13200) (Show Source): You can put this solution on YOUR website!
As the other tutor indicates, the method for solving all three of your problems is the same, so I will only address the first.
The response from that other tutor shows a standard solution method taught in virtually all references. With that method, we find the x values where the function value is 0 and divide the x axis into intervals using those x values; then we use a test point in each interval to find the interval(s) in which the given inequality is satisfied.
For a quadratic inequality, with two zeros creating 3 intervals, that process is relatively efficient. But for higher degree polynomials, or for rational functions, where the number of intervals might be much greater than 3, there is a more efficient way to determine the correct intervals.
This more efficient way for finding the correct intervals uses the fact that, as we "walk" along the x axis, the sign of the function can change only at the zeros of the function (or, in the case of rational functions, the zeros of either the numerator or denominator).
It is usually easiest to start to the right of every zero, where all factors of the function are positive, and move to the left to see where the sign of the function changes.
For your first example....
The zeros are at x=1 and x=4.
So pick a value greater than 4 and see that the function value is positive.
Then, as you "walk" along the number line to the left, the function value changes sign each time you pass one of the roots. So the function value is negative between 1 and 4, and positive again for x less than 1.
ANSWER: The inequality is satisfied on (-infinity,1) and (4,infinity).
Of course, a completely different solution method, once the roots x=1 and x=4 are determined, is to know that the graph is an upward-opening parabola, so the inequality is NOT satisfied only between the two roots.
Answer by josgarithmetic(39618) (Show Source): You can put this solution on YOUR website!
The second inequality
Important values to check around on number line are +1 and +2.
To left of k=1
(0-2)(0-1)
(-2)(-1)<0 FALSE
Between the two values
(1.5-2)(1.5-1)
(-)(+)<0 TRUE
To Right of k=2
(3-2)(3-1)
(+)(+)<0 FALSE
SOLUTION:
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