Solve inequation

A

acemi123

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Consider that, Q(x) is a second-degree polinom. Which one is true and why explain, please?

a) just A and C have the same solutions
b) Three inequations have the same solutions
c) Just B and C have the same solutions
d) Just A and B has the same solutions
 
Hi. Can you please tell us where you are stuck? Maybe show us what you tried?

Here are some questions that should help you. What is the minimum value of x^2 + 3? Can x^2 - 3x + 9 be both positive and negative?
 
Jomo said:
Hi. Can you please tell us where you are stuck? Maybe show us what you tried?

Here are some questions that should help you. What is the minimum value of x^2 + 3? Can x^2 - 3x + 9 be both positive and negative?
Click to expand...
Q(x) is positive so I think so the denominator of B also has to be positive, same thing for C, but not knowing Q(x) how to find all solutions? So which one is true?
 
acemi123 said:
I think there are no real roots for B.
b^-4ac > 0 but (-3)^ -4.1.9 = -27 its smaller than 0 . It means no real root. Am I right?
Click to expand...
OK, good work. So x^2- 3x+ 9 has no real roots. What does that mean for this problem?
 
acemi123 said:
View attachment 11784
Consider that, Q(x) is a second-degree polynomial. Which one is true and why explain, please?
b) Three inequations have the same solutions
Click to expand...
I do not think that you correctly understand what is being asked of you.
Given the three inequality in which \(\displaystyle Q(x)\) is is a second-degree polynomial what are possible solutions.
Look \(\displaystyle A)~\frac{Q(x)}{1}>0,~\;B)~\frac{Q(x)}{\left(x^2-\tfrac{3}{2}\right)^2+\tfrac{27}{4}}>0,~\;\&~\;C)\frac{Q(x)}{x^2+3}>0\)
The denominator of each of those three fractions is positive for any value of \(\displaystyle x\).
The polynomial \(\displaystyle Q(x)\) is the numerator and is the only determinate of the sign for each fraction.
Do you now see why b) The three inequalities have the same solutions is the only correct answer?
 
acemi123 said:
I think there are no real roots for B.
b^-4ac > 0 but (-3)^ -4.1.9 = -27 its smaller than 0 . It means no real root. Am I right?
Click to expand...
So this denominator is always positive or always negative. If you plug in any value for x you will get a positive value. So it is always positive.
In fact, as pka pointed out, all the denominators are positive. So the factions will be positive only when the numerator, which happens to all be Q(x), is positive.
 
pka said:
I do not think that you correctly understand what is being asked of you.
Given the three inequality in which \(\displaystyle Q(x)\) is is a second-degree polynomial what are possible solutions.
Look \(\displaystyle A)~\frac{Q(x)}{1}>0,~\;B)~\frac{Q(x)}{\left(x^2-\tfrac{3}{2}\right)^2+\tfrac{27}{4}}>0,~\;\&~\;C)\frac{Q(x)}{x^2+3}>0\)
The denominator of each of those three fractions is positive for any value of \(\displaystyle x\).
The polynomial \(\displaystyle Q(x)\) is the numerator and is the only determinate of the sign for each fraction.
Do you now see why b) The three inequalities have the same solutions is the only correct answer?
Click to expand...
It is the correct answer, thank you but to be sure I found another one. If the solution depends on P(x) why this one has a different solution?
I will tell you the answer after your explaining for bottom inequation.

11807


a) just A and C have the same solutions
b) Three inequations have the same solutions
c) Just B and C have the same solutions
d) Just A and B have the same solutions
 

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You should know that the polynomial \(\displaystyle px^2+qx+t=0,~p\ne 0\) has two real roots if and only if \(\displaystyle q^2-4pt>0\).
That is, its discriminate is positive.
 
acemi123 said:
It is the correct answer, thank you but to be sure I found another one. If the solution depends on P(x) why this one has a different solution?
I will tell you the answer after your explaining for bottom inequation.

View attachment 11807


a) just A and C have the same solutions
b) Three inequations have the same solutions
c) Just B and C have the same solutions
d) Just A and B have the same solutions
Click to expand...
Have you pondered how to solve this, or just looked up the answer? You have to do the former in order to learn. And if you did, you should see the difference between the problems.

Each of the inequalities will be true whenever P(x) > 0 and the denominator is positive, OR when P(x) < 0 and the denominator is negative.

In the original problem, all the denominators were always positive, so the solutions were exactly the solutions of P(x) > 0.

In this problem, one of the denominators is sometimes negative. Specifically, x^4 - 1 can be negative, whereas x^4 + 1 is always positive. Can you see why this is so?

As a result, (B) will have a different truth value in (-1, 1). So the answer is that only A and C have the same solution.

Did you come up with that reasoning? It's exactly the same reasoning you were shown for the first problem, but with a different conclusion because of the different denominators.
 
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