# Inequality problem

1. Jul 2, 2013

### Pranav-Arora

1. The problem statement, all variables and given/known data
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.

2. Relevant equations

3. 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
$$x_1=\frac{a+\sqrt{2(a-1)}}{2}, x_2=\frac{a-\sqrt{2(a-1)}}{2}$$
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.
$$4(2)^2-4a(2)+a^2-2a+2=3$$
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!

Last edited: Jul 2, 2013
2. Jul 2, 2013

### tiny-tim

Hi Pranav-Arora!

You seem to be misinterpreting the question.

3. Jul 2, 2013

### Pranav-Arora

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. Jul 2, 2013

### tiny-tim

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

5. Jul 2, 2013

### Pranav-Arora

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.

6. Jul 2, 2013

yes

7. Jul 2, 2013

### Pranav-Arora

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?

8. Jul 2, 2013

### tiny-tim

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

9. Jul 2, 2013

### Pranav-Arora

$a^2-2(a-1)$

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

10. Jul 2, 2013

### tiny-tim

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

11. Jul 2, 2013

### 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.

12. Jul 2, 2013

### Pranav-Arora

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. Jul 2, 2013

### haruspex

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. Jul 4, 2013

### Pranav-Arora

Bump...

15. Jul 5, 2013

### 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?