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

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Question Stats: 35% (01:54) correct 65% (02:20) wrong based on 182 sessions

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

(1) $$a + b = 4$$
(2) $$a^2 + b^2 = 8$$
Math Expert V
Joined: 02 Sep 2009
Posts: 58464
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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.

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

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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.

Intern  B
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Re: Is a^4+b^4 less than 32?  [#permalink]

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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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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

For more check Ultimate GMAT Quantitative Megathread

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

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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, $$2(a^4+b^4)>=64$$, $$(a^4+b^4)>=32$$,
Intern  B
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Is a^4+b^4 less than 32?  [#permalink]

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

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_________________ Re: Is a^4+b^4 less than 32?   [#permalink] 07 Feb 2019, 01:43
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