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# If a and b represent positive real numbers and (a^4+b^2)^2=198 and a^8

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Joined: 02 Sep 2009
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If a and b represent positive real numbers and (a^4+b^2)^2=198 and a^8  [#permalink]

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02 Dec 2019, 23:37
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Difficulty:

35% (medium)

Question Stats:

76% (01:41) correct 24% (02:34) wrong based on 46 sessions

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Competition Mode Question

If a and b represent positive real numbers and $$(a^4+b^2)^2=198$$ and $$a^8+b^4=36$$, what is the value of $$a\sqrt{b}$$?

A. 2.6
B. 3.0
C. 3.4
D. 4.0
E. 4.2

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Re: If a and b represent positive real numbers and (a^4+b^2)^2=198 and a^8  [#permalink]

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03 Dec 2019, 01:38
1
from above equations we get
a^4*b^2=81
a(sqrtb)= 3

OA:B
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Re: If a and b represent positive real numbers and (a^4+b^2)^2=198 and a^8  [#permalink]

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03 Dec 2019, 02:02
1
simplify

(a4+b2)2=198(a4+b2)2=198
a8+b^4+2a^4*b^2 = 198
given
a8+b4=36

2a^4*b^2 = 198-36
a^4*b^2= 81
square both sides
a^2*b=9
again square
a√b=3
IMO B

If a and b represent positive real numbers and (a4+b2)2=198(a4+b2)2=198 and a8+b4=36, what is the value of ab‾√ab?

A. 2.6
B. 3.0
C. 3.4
D. 4.0
E. 4.2
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Re: If a and b represent positive real numbers and (a^4+b^2)^2=198 and a^8  [#permalink]

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03 Dec 2019, 04:36
1
$$(a^4+b^2)^2=198$$
$$a^8+b^4$$$$+2a^4.b^2=198$$
$$36$$$$+2a^4.b^2=198$$
$$2a^4.b^2=162$$ --> $$a^4.b^2=81$$
$$(a√b)^4=3^4$$ --> $${a√b}=3$$, because a and b are positive real numbers.

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Re: If a and b represent positive real numbers and (a^4+b^2)^2=198 and a^8  [#permalink]

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03 Dec 2019, 04:49
1
Quote:
If $$a$$ and $$b$$ represent positive real numbers and $$(a^4+b^2)^2=198$$ and $$a^8+b^4=36$$, what is the value of $$a\sqrt{b}$$?

A. 2.6
B. 3.0
C. 3.4
D. 4.0
E. 4.2

$$a^8+b^4=36$$
$$(a^4+b^2)^2=198…a^8+b^4+2a^4b^2=198$$
$$(36)+2a^4b^2=198…a^4b^2=81…a^2b=9…a\sqrt{b}=3$$

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Re: If a and b represent positive real numbers and (a^4+b^2)^2=198 and a^8  [#permalink]

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03 Dec 2019, 06:35
1
If a and b represent positive real numbers and $$(a^4+b^2)^2=198$$ and $$a^8+b^4=36$$, what is the value of a√b?

A. 2.6
B. 3.0
C. 3.4
D. 4.0
E. 4.2

$$(a^4 + b^2)^2 = 198$$
$$a^8+b^4 + 2.a^4.b^2 = 198$$
$$36 + 2.a^4.b^2 = 198$$
$$a^4.b^2 = 81$$
$$a^2.b = 9$$
$$a√b = - 3$$ or $$3$$

Since a is positive and √b can't be negative, answer is 3.

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Re: If a and b represent positive real numbers and (a^4+b^2)^2=198 and a^8  [#permalink]

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03 Dec 2019, 12:44
1
Given that (a^4+b^2)^2=198 and a^8+b^4=36 we are to determine a√b.
(a^4+b^2)^2=a^8+2a^4b^2+b^4=198
but a^8+b^4=36
hence 36+2a^4b^2=198
2a^4b^2=162
a^4b^2=81
a^2b=9
a√b=3

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Re: If a and b represent positive real numbers and (a^4+b^2)^2=198 and a^8  [#permalink]

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03 Dec 2019, 15:32
1
a, b —real positive numbers

$$(a^{4} +b^{2})^{2} = 198$$

$$a^{8} +b^{4} = 36$$

—> $$a*b^{1/2}$$ = ???

$$a^{8} + 2a^{4}b^{2} + b^{4}= 198$$

$$a^{8} + b^{4}= 36$$

—> $$2a^{4}b^{2}= 162$$

$$a^{4}b^{2}= 81$$

—>$$(ab^{1/2})^{4}= 81$$

$$ab^{1/2}= 3$$

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Re: If a and b represent positive real numbers and (a^4+b^2)^2=198 and a^8  [#permalink]

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05 Dec 2019, 20:23
1
Bunuel wrote:

Competition Mode Question

If a and b represent positive real numbers and $$(a^4+b^2)^2=198$$ and $$a^8+b^4=36$$, what is the value of $$a\sqrt{b}$$?

A. 2.6
B. 3.0
C. 3.4
D. 4.0
E. 4.2

Foiling the first equation, we have:

a^8 + b^4 + 2(a^4)(b^2) = 198

Since a^8 + b^4 = 36, we have:

36 + 2(a^4)(b^2) = 198

(a^4)(b^2) = 81

a^2 * b = 9

a√b = 3

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Re: If a and b represent positive real numbers and (a^4+b^2)^2=198 and a^8   [#permalink] 05 Dec 2019, 20:23
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