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Is a^4+b^4 less than 32?

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Is a^4+b^4 less than 32?  [#permalink]

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New post 08 Dec 2017, 21:42
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Is \(a^4+b^4\) less than 32?

(1) \(a + b = 4\)
(2) \(a^2 + b^2 = 8\)
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Is a^4+b^4 less than 32?  [#permalink]

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New post 08 Dec 2017, 23:35
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Is \(a^4+b^4\) less than 32?

You can solve this problem by testing values or use the following properties:

For a given sum of two numbers, the sum of even powers of these numbers is minimized, when they are equal. Simply put if it's given that \(x + y = 10\), then \(x^{(even \ integer)} + y^{(even \ integer)}\) is minimum when \(x = y = 5\). So, the minimum value of \(x^2 + y^2\), given that \(x + y = 10\) is \(5^2 + 5^2 = 50\).

For a given sum of two numbers their product is maximized when they are equal. Simply put if it's given that \(x + y = 10\), then xy is maximized when \(x = y = 5\). So, the maximum value of xy, given that \(x + y = 10\) is 5*5 = 25.



(1) \(a + b = 4\).

According to the first property above. \(a^4 + b^4\) will be minimized given that \(a + b = 4\), when \(a = b = 2\). In this case \(a^4 + b^4 = 2^4 + 2^4 = 32\). So, the minumum possible value of \(a^4 + b^4\) is 32 (not less than 32). Sufficient.


(2) \(a^2 + b^2 = 8\)

You can use the same property (which would be easier) or square and test the second property:
\(a^4 + 2a^2b^2 + b^4 = 64\)
\(a^4 + b^4 = 64 -2a^2b^2\)

According to the second property \(2a^2b^2\) is maximized given that \(a^2 + b^2 = 8\), when \(a^2 = b^2 = 4\). So, the maximum value of \(2a^2b^2\) is 32. This makes the minimum value of \(a^4 + b^4 = 64 -2a^2b^2\) equal to 64 - 32 = 32. So, the minumum possible value of \(a^4 + b^4\) is 32 (not less than 32). Sufficient.

Answer: D.

hope it's clear.
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Re: Is a^4+b^4 less than 32?  [#permalink]

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New post 08 Dec 2017, 23:39
diota2004 wrote:
Is \(a^4+b^4\) less than 32?

(1) \(a + b = 4\)
(2) \(a^2 + b^2 = 8\)



(1) a + b = 4
We can try using some plugging in of numbers. First thing that comes to mind is 'what if a & b are each equal?'. So when we put a=b=2, we get a^4 + b^4 = 16+16=32.
Now if we give different values to them, eg, put a=3, b=1, we have a^4 + b^4 = 81+1 =82, or if we put a=0, b=4, we have a^4 + b^4 = 0+256 =256. Even if we give negative value to one of a or b, it wont matter because powers are even, it will convert to positive only. eg, if a=-1, b=5, we have a^4 + b^4 =1+625=626

So this plugging of numbers tells me that no matter what values to a&b I give, if a+b = 4, then a^4 + b^4 will always be greater than or equal to 32 (least value 32 will occur when a & b are equal). so a^4 + b^4 can never be less than 32. Sufficient


(2) a^2 + b^2 = 8
Note that a^4 = (a^2)^2 and b^4 = (b^2)^2. So we can proceed with same plugging in of numbers. First lets put a^2=b^2 = 4. In this case a^4 + b^4 = 4^2 + 4^2 = 16+16=32. Now if we substitute unequal values of a^2/b^2, eg, a^2 = 7, b^2 = 1, we get a^4 + b^4 = 49+1=50. Again, no matter which values we give to a^2 and b^2, as long as a^2+b^2 = 8, we will always have a^4 + b^4 as greater than or equal to 32 (least value 32 will occur when a^2 & b^2 are equal). So a^4 + b^4 can never be less than 32. Sufficient.


Hence D answer
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Re: Is a^4+b^4 less than 32?  [#permalink]

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New post 11 Jan 2018, 01:15
I got the answer correct based on the following thoughts:
a^4 + b^4 = (a+b)^4 < 32?

1) Since I know the value of (a+b)=4 I know that I will get a value for (a+b)^4 and therefore will be able to answer the question (I don't need to plug numbers and calculate) -> suff.
2.) a^2+a^2= (a+b)^2 = 8. Same reasoning as above, I habe a value for the term, and since (a+b)^4 = ((a+b)^2)^2 I know that I am able to calculate a value and therefore it is sufficient (no need to calcuate value) -> suff.

Does this make sense or did I miss something and was perhaps just lucky?
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Re: Is a^4+b^4 less than 32?  [#permalink]

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New post 11 Jan 2018, 01:24
Gregsterh wrote:
I got the answer correct based on the following thoughts:
a^4 + b^4 = (a+b)^4 < 32?

1) Since I know the value of (a+b)=4 I know that I will get a value for (a+b)^4 and therefore will be able to answer the question (I don't need to plug numbers and calculate) -> suff.
2.) a^2+a^2= (a+b)^2 = 8. Same reasoning as above, I habe a value for the term, and since (a+b)^4 = ((a+b)^2)^2 I know that I am able to calculate a value and therefore it is sufficient (no need to calcuate value) -> suff.

Does this make sense or did I miss something and was perhaps just lucky?


The above is wrong.

Generally \(a^4 + b^4 \neq (a+b)^4\) and \(a^2 + b^2 \neq (a+b)^2\). Does \(1^2 + 2^2 = (1+2)^2\)?

\((a+b)^2=a^2+2ab+b^2\) NOT \(a^2 + b^2\).

7. Algebra



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Hope it helps.
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Is a^4+b^4 less than 32?  [#permalink]

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New post 14 Jan 2018, 21:53
Condition 1: a+b=4, and \((a+b)^2=16\), hence \(a^2+b^2+2ab=16\), therefore \(2ab=16-(a^2+b^2)\)
also we know, \(a^2+b^2>=2ab\)
Now put the value of 2ab in inequality
\(a^2+b^2>=16-(a^2+b^2)\)
\(2(a^2+b^2)>=16\), hence \((a^2+b^2)>=8\), therefore, \((a^2+b^2)^2>=64\), \(a^4+b^4+2a^2b^2>=64\)
again, \(a^4+b^4>=2a^2b^2\)
with two equalities we get
\(2(a^4+b^4)>=64\),
\(a^4+b^4>=32\)
Condition 2: \((a^2+b^2)=8\), \((a^2+b^2)^2=64\), \(a^4+b^4+2a^2b^2=64\)
also, \(a^4+b^4>=2a^2b^2\)
this implies, \(2(a^4+b^4)>=64\), \((a^4+b^4)>=32\),
Hence answer D
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Is a^4+b^4 less than 32?  [#permalink]

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New post 21 Jan 2018, 01:43
Another way:

Statement (1)

\(a = 2 – m\)
\(b = 2 + m\)

\(a^4+b^4 = (2-m)^4 + (2+m)^4 = 32 + 2m^4 + 48m^2 >= 32\)
\(a^4+b^4\) cannot be less than 32.
->Sufficient

Statement (2)

\(a^2 = 4 – m\)
\(b^2 = 4 + m\)

\(a^4 + b^4 = (4-m)^2 + (4+m)^2 = 32 + 2m^2 >= 32\)
\(a^4+b^4\) cannot be less than 32.
->Sufficient

The answer is D
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