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

[Saitama]

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}##.

Any help is appreciated. Thanks!

 

[tiny-tim]

Hi Pranav-Arora!

You seem to be misinterpreting the question.

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

 

[Saitama]

Hi Pranav-Arora!

You seem to be misinterpreting the question.

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

I understand that the question asks me about the minimum.

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? :)

 

[tiny-tim]

i'd start by rewriting the original equation in the form (x - b)² - c² ≥ 0

 

[Saitama]

i'd start by rewriting the original equation in the form (x - b)² - c² ≥ 0

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.

 

[tiny-tim]

yes

 

[Saitama]

yes

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?

 

[tiny-tim]

I came up with this: ##(2x-a)^2-2(a-1)##

and the minimum of that (for 0 ≤ x ≤ 2) is … ?

And how did you get that greater than or equal to symbol?

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! ​

 

[Saitama]

and the minimum of that (for 0 ≤ x ≤ 2) is … ?

##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! ​

Why greater than or equal to zero? Why not greater than or equal to 3? I guess I am missing something really basic.

 

[tiny-tim]

Why not greater than or equal to 3?

i preferred to get everything over onto the LHS (and only 0 on the RHS)

 

[haruspex]

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.

 

[Saitama]

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.

 

[haruspex]

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.

 

[Saitama]

Bump...

 

[verty]

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?