Solving Inequality Problem: Find a for 3 on 0-2 Interval

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    Inequality
In summary: I had ##(2x-a)^2-2(a-1)##. The squared term is zero when x=a/2. Case I:##0 \leq a/2 \leq 2 \Rightarrow 0 \leq a \leq 4##For this case, the minimum value is at a/2 i.e ##-2(a-1)##. This is equal to 3 when a=-1/2 but this is not possible as a ranges between 0 and 4.Case II:When ##a/2 > 2 \Rightarrow a > 4##, the minimum value is at 2.Solving
  • #1
Saitama
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Homework Statement


Find all numbers ##a## for each of which the least value of the quadratic trinomial ##4x^2-4ax+a^2-2a+2## on the interval ##0\leq x \leq 2## is equal to 3.

Homework Equations


The Attempt at a Solution


I don't really know what should be the best way to start with this type of question. I started with finding out the roots of the given equation. The roots come out to be
[tex]x_1=\frac{a+\sqrt{2(a-1)}}{2}, x_2=\frac{a-\sqrt{2(a-1)}}{2}[/tex]
The roots exist when ##a>1## and it is easy to see that ##x_1>x_2## for ##a>1##. I don't know how to proceed ahead. 0 and 2 should not lie between the roots as we need the minimum value to be 3. Also, ##x_1## is always greater than zero for ##a>1## so 0 and 2 should lie before ##x_1## and ##x_2## on a number line. Hence, the least value will be at ##x=2## and this should be equal to 3.
[tex]4(2)^2-4a(2)+a^2-2a+2=3[/tex]
Solving this, I get ##a=5+\sqrt{10}, 5-\sqrt{10}##. Both the values are greater than hence both are admissible but the answer key does not show up the second value i.e ##5-\sqrt{10}##. :confused:

Any help is appreciated. Thanks!
 
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  • #2
Hi Pranav-Arora! :smile:

You seem to be misinterpreting the question.

The question isn't about the roots, it's about the minimum. :confused:
 
  • #3
tiny-tim said:
Hi Pranav-Arora! :smile:

You seem to be misinterpreting the question.

The question isn't about the roots, it's about the minimum. :confused:

I understand that the question asks me about the minimum. :smile:

0 and 2 should not lie between the roots. The given minimum is 3 which is a positive value, and between the roots, the value of quadratic is negative.

But I am still not sure about how should I begin? What I posted were my thoughts about approaching the problem but I feel that is incorrect. Can I have a few hints? :)
 
  • #4
i'd start by rewriting the original equation in the form (x - b)2 - c2 ≥ 0 :wink:
 
  • #5
tiny-tim said:
i'd start by rewriting the original equation in the form (x - b)2 - c2 ≥ 0 :wink:

Sorry tiny-tim, there was a typo in the first post, I have corrected it. The correct quadratic is ##4x^2-4ax+a^2-2a+2##, I accidentally wrote -2x instead of -2a. Should I still follow your hint? Sorry for the trouble. :redface:
 
  • #6
yes :smile:
 
  • #7
tiny-tim said:
yes :smile:

I came up with this: ##(2x-a)^2-2(a-1)## but this doesn't match up with what you asked me. And how did you get that greater than or equal to symbol? :confused:
 
  • #8
Pranav-Arora said:
I came up with this: ##(2x-a)^2-2(a-1)##

and the minimum of that (for 0 ≤ x ≤ 2) is … ? :smile:
And how did you get that greater than or equal to symbol? :confused:

well, we want a minimum, and the question is posed as an inequality problem …

so i thought it was time to throw in an inequality! :biggrin:
 
  • #9
tiny-tim said:
and the minimum of that (for 0 ≤ x ≤ 2) is … ? :smile:
##a^2-2(a-1)##

well, we want a minimum, and the question is posed as an inequality problem …

so i thought it was time to throw in an inequality! :biggrin:

:confused:
Why greater than or equal to zero? Why not greater than or equal to 3? I guess I am missing something really basic. :rolleyes:
 
  • #10
Pranav-Arora said:
Why not greater than or equal to 3?

i preferred to get everything over onto the LHS (and only 0 on the RHS) :smile:
 
  • #11
At some stage I think you're going to have to break it into cases. You're looking for the minimum only within the range [0, 2]. In the OP you considered the possibility that the minimum occurs at x=2. You found two values for a that give f(2)=3 , but you did not check whether either of these give f(x)>=3 for all x in [0, 2].
You also seem to have ruled out other locations for the minimum erroneously. There are two cases to consider for alternative locations of the minimum.
 
  • #12
tiny-tim, I still do not understand your inequality method but meanwhile, I tried something else and reached the answer but not sure if my approach is correct.

I had ##(2x-a)^2-2(a-1)##. The squared term is zero when x=a/2.

Case I:
##0 \leq a/2 \leq 2 \Rightarrow 0 \leq a \leq 4##
For this case, the minimum value is at a/2 i.e ##-2(a-1)##. This is equal to 3 when a=-1/2 but this is not possible as a ranges between 0 and 4.

Case II:
When ##a/2 > 2 \Rightarrow a > 4##, the minimum value is at 2.
Solving the quadratic, I get ##a=5+\sqrt{10}, 5-\sqrt{10}##. Second value is not possible as a is greater than 4.

Case III:
When ##a/2 < 0 \Rightarrow a<0##, the minimum value is at 0. From this I get two values ##a=1-\sqrt{2}, 1+\sqrt{2}##. Only the first value is possible.

Hence, ##a=1-\sqrt{2}, 5+\sqrt{10}##. This is the correct answer however I am unsure about my approach.

I am still interested in understanding tiny-tim's method.
 
  • #13
Pranav-Arora said:
tiny-tim, I still do not understand your inequality method but meanwhile, I tried something else and reached the answer but not sure if my approach is correct.

I had ##(2x-a)^2-2(a-1)##. The squared term is zero when x=a/2.

Case I:
##0 \leq a/2 \leq 2 \Rightarrow 0 \leq a \leq 4##
For this case, the minimum value is at a/2 i.e ##-2(a-1)##. This is equal to 3 when a=-1/2 but this is not possible as a ranges between 0 and 4.

Case II:
When ##a/2 > 2 \Rightarrow a > 4##, the minimum value is at 2.
Solving the quadratic, I get ##a=5+\sqrt{10}, 5-\sqrt{10}##. Second value is not possible as a is greater than 4.

Case III:
When ##a/2 < 0 \Rightarrow a<0##, the minimum value is at 0. From this I get two values ##a=1-\sqrt{2}, 1+\sqrt{2}##. Only the first value is possible.

Hence, ##a=1-\sqrt{2}, 5+\sqrt{10}##. This is the correct answer however I am unsure about my approach.

I am still interested in understanding tiny-tim's method.
That looks good. I broke it into cases according to whether the min was at 0, 2, or in-between, but the two methods are pretty much equivalent. I doubt tiny-tim's method was much different either, but it would be interesting to see.
 
  • #14
Bump...
 
  • #15
I believe Tiny-tim was using calculus to get the answer because he had you put it in the best form to apply calculus to solve it. I won't give that method though. As a precalculus question, using the axis of symmetry is the most intuitive way, I believe. And Tim's final formula is the turning point form, so he might well have been hinting that you think about the axis of symmetry or use calculus, which I think is absolutely correct.

Okay?
 

FAQ: Solving Inequality Problem: Find a for 3 on 0-2 Interval

How do you define an inequality problem?

An inequality problem is a mathematical equation that contains an inequality symbol, such as <, >, ≤, or ≥. It involves finding the values of the variable that make the inequality true.

What is the purpose of solving inequality problems?

The purpose of solving inequality problems is to find the set of values that satisfy the given inequality. This helps in understanding the relationship between different variables and making decisions based on the given constraints.

How do you solve an inequality problem?

To solve an inequality problem, you need to follow the same rules as solving an equation, such as simplifying both sides, combining like terms, and isolating the variable. However, when multiplying or dividing both sides by a negative number, the direction of the inequality symbol must be reversed.

What is the meaning of "a for 3 on 0-2 Interval" in an inequality problem?

The phrase "a for 3 on 0-2 Interval" means that the unknown variable, represented by the letter "a", is the solution to the inequality when it is restricted within the interval of 0 and 2. This means that the value of "a" must be greater than or equal to 0 and less than or equal to 2.

Can you provide an example of solving an inequality problem with the given parameters?

For the inequality problem "2x + 3 < 7", the solution would be "a for 3 on 0-2 Interval". This means that "a" can take on any value between 0 and 2 to make the inequality true. Solving the inequality, we get x < 2, which can be represented as "a for 2 on 0-2 Interval".

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