A minimum integer value
As x ranges over all real values, what is the smallest integer value of
1 1 x 2 + 2 2 x + 3 3 3 ?
Details and assumptions
Note that we are not asking for the value of x .
The answer is 20.
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3 solutions
Moderator note:
Make sure to explain your steps clearly, so that others can understand it. If you are using a certain approach, mentioning it helps others to immediately associate what you are doing, instead of taking that it's magic. For example,
By Completing the square , we get that 1 1 x 2 + 2 2 x + 3 3 3 = 1 1 ( x + 1 1 ) 2 + 2 1 2
Hi I will try to elaborate on his explanation, hopefully it helps :)
So after simplifying the equation to ((x+11)^2+212)/11, we can now see that the whole term would have to be divisible by 11 for the answer to be an integer. The closest term to 212 that can be divisible by 11 and is bigger than 212 is 220, therefore showing that the smallest integer is 20. This would also mean that (x+11)^2 = 8 and you can then work out x, But the value of x is irrelevant as we want the overall value of the equation.
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ah darn it, "what is the smallest integer value of the equation" and I have been trying to find the smallest integer value of x. Thanks for making me read the question again :)
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I'd like to point out that the exact value of x is not important in this question.
Smallest integer value of x also won't make much sense, in this question.
I solved the problem this way...
The equation x 2 + 2 2 x + 3 3 3 does not yield any real solution... Since x is a real number, and we need to find the smallest integer value of 1 1 x 2 + 2 2 x + 3 3 3 , we will divide the terms in two parts... One of the parts will be a polynomial with real solution with the lowest possible discriminant... Let this polynomial be x 2 + 2 2 x + c ... And the other part will be a constant and the remainder which is a multiple of 1 1 , let that be m ... This will bring us the lowest possible integer value as the discriminant to be the lowest, the value of c should be the highest leaving the lowest possible multiple of 1 1 ... The structure should look like this...
1 1 x 2 + 2 2 x + c + 1 1 m [ c + m = 3 3 3 ]
Let's move on... The discriminant of the quadratic x 2 + 2 2 x + c = 0 is 2 2 2 − 4 c = 4 8 4 − 4 c ... Here, 4 8 4 − 4 c ≥ 0 ⇒ c ≤ 1 2 1
Trying c = 1 2 1 , we find that the remainder m = 3 3 3 − 1 2 1 = 2 1 2 is not a multiple of 1 1 ... Hence, c is less than 1 2 1 and the remainder m is more than 2 1 2 ... Since, ⌊ 1 1 2 1 2 ⌋ = 1 9 , we can try the next integer 2 0 and it works... Now the remainder is m = 2 0 × 1 1 = 2 2 0 and the value of c is 3 3 3 − 2 2 0 = 1 1 3 ... Now the discriminant is 2 2 2 − ( 4 × 1 1 3 ) = 4 8 4 − 4 5 2 = 3 2 , which is the lowest possible discriminant that yields real solutions for x ... Now the overall is...
1 1 x 2 + 2 2 x + 1 1 3 + 1 1 2 2 0
Now, there exists a real solution of x for the equation x 2 + 2 2 x + 1 1 3 = 0 ... Whatever the solution would be, let's cancel out the first part of our overall look-up and the result is 1 1 2 2 0 = 2 0
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Its a simple maxima minima problem. Just differentiate the above term and put it as equal to zero. d/dx(x^2+22x+333/11)= 2x+22=0=> x=-11. put x = -11 in the equation u will get 212/11.. Round off the number to integer and u will get 20.
Can you please elaborate your solution, are you saying the value of x is 8 ?
i dont understand
i really dont understand
What is the correct answer? How come that (x^2+22x+333)11 = [(x+11)^2+212]/11? Where is now the "333"? Kindly elaborate. And what is x equal to?
havn't got the faintest idea on how to do this...
So, what is the actual answer to the question? I've read through what everyone has written, yet I don't seem to get to the conclusion of what the answer is?!?!?!
okay, i didn't want to buy a clue to how to simplify but as to more or less find the way to submit the solution i come up with not being able to find the way to square, cube, or properly provide all necessary components in my equations solution or for typing out formula.
edit menu Yash T. Yash T. I understand the logic of how you guys arrived at 20 but can anyone tell me what is the actual value of x? It seems that x can only be an integer . Also I did not find any integer's square up to 10 which is congruent to 8(mod 11). There seems to be no value which will satisfy the above condition which is that x^2 congruent to 8(mod 11). Can anyone enlighten me?
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You do not need to know the value of x since the question is not asking for it. Furthermore, x is not an integer.
but how is (x+11) squared divisible by 11, to give 8?
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er no is the whole equation must be divisible by 11, so (x+11) squared must be equals to 8 to give the smallest overall integer that can be divisible by 11
Have you tried with all the numbers even 8?
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alguem pode falar em portugues por favor?????????????????????
i don't spiking english
let us say x^2+22x+333=a =>x^2+22x+333-a=0 so this is a quadratic equation the roots of this equation will be real if 22^2 - 4*(333-a) >=0 solving this we will get a>=212; so in the given problem if values of x are real then the numerator has to be greater than 212.. the nearest multiple of 11 greater than 212 is 220... and the ans is 220/11=20
I understand the logic of how you guys arrived at 20 but can anyone tell me what is the actual value of x? It seems that x can only be an integer only. Also I did not find any integer's square up to 10 which is congruent to 8(mod 11). There seems to be no value which will satisfy the above condition which is that x^2 congruent to 8(mod 11). Can anyone enlighten me?
The logic of your calculations I understand. But shouldn't x be number so that x^2 +3 should be divisible by 11 and is there any value of x that satisfies it. No!
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x does not have to be an integer, just real. So the value of x is the square root of 5. However, the value of x is irrelevant in this question
The given equation y = 1 1 x 2 + 2 2 x + 3 3 3 is the equation of a parabola which opens upward. Consider the quadratic equation y = a x 2 + b x + c . This equation can be simplified as follows, y = a x 2 + b x + c = a ( x 2 + a b x + a c ) . Using completing the square method we can simplify the equation to y = a ( x 2 + 2 a 2 b x + 4 a b 2 − 4 a b 2 + a c ) ⇒ y = a ( ( x + 2 a b ) 2 − 4 a b 2 − 4 a c ) . Now it is easy to understand that when a > 0 , y attains minimum when x = 2 a − b → y = 4 a − ( b 2 − 4 a c ) , similarly when a < 0 , y attains maximum at x = 2 a − b → y = 4 a − ( b 2 − 4 a c ) . Thus from this lengthy (unneeded) proof I have proved that our given quadratic equation will be minimum at y m i n = 1 1 4 − ( 2 2 − 4 ∗ 1 1 1 ∗ 1 1 3 3 3 ) = 1 9 . 2 7 . Thus the minimum integer value near the quadratic is 2 0 .
Althought I used that method, I didn't know how was its proof so thanks for your explanation!
(x^2+22x+333)/11=((x+11)^2+212)/11
because ((x+11)^2+212)/11 is integrer,((x+11)^2+212) must can be divided by 11, the minimal possible of (x+11)^2 to be divided by 11 is 8
so (8+212)/11=20