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If a and b are negative and a

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Manager
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S
Joined: 31 Oct 2018
Posts: 77
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If a and b are negative and a  [#permalink]

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New post 18 Apr 2019, 09:35
4
00:00
A
B
C
D
E

Difficulty:

  55% (hard)

Question Stats:

61% (02:39) correct 39% (02:29) wrong based on 44 sessions

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If a and b are negative and \(a^2b^2\) = 21-4ab then \(a^2\) = ?

A. (16-4b)/\(b^3\)

B. 28/\(b^3\)

C. 16/(\(b^2\)+7b)

D. 49/\(b^2\)

E. 9/\(b^2\)

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Manager
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Re: If a and b are negative and a  [#permalink]

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New post 18 Apr 2019, 12:07
1
\(a^2b^2 = 21 - 4ab\)
\(a^2b^2 + 4ab = 21\)
\(ab(ab + 4) = 21\)

Since 21 = 7 * 3, \(ab\) must be \(3\) in order for equation to hold true.
\(a^2 = \frac{21 - 4ab}{b^2} = \frac{9}{b^2}\). Pick E.
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If a and b are negative and a  [#permalink]

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New post 18 Apr 2019, 13:23
Hi!

\(a^2b^2\) = 21-4ab

\(\frac{9}{b^2}b^2 + 4\frac{3}{b}b = 21\)

\(9 + 12 = 21\)

E
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Re: If a and b are negative and a  [#permalink]

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New post 18 Apr 2019, 15:28
1
In the question we are given

a^2b^2=21-4ab
=>a^2b^2+4ab-21=0

Lets try to convert this into the form (x+y)^2=x^2+2xy+y^2

a^2b^2+4ab-21=0
or a^2b^2+4ab=21

Add 4 to both sides, we get

a^2b^2+4ab+4=21+4

=>(ab)^2+4ab+4=25

=>(ab+2)^2=25

=>(ab+2)^2=5^2

=>ab+2=5

=>ab=3

=>a=3/b

=>a^2=9/b^2

Hence option E
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Re: If a and b are negative and a  [#permalink]

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New post 21 Apr 2019, 11:45
For me it is between D and E.
Substituting D and E for a^2 makes LHS = RHS.
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If a and b are negative and a  [#permalink]

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New post 21 Apr 2019, 12:15
1
btrg wrote:
If a and b are negative and \(a^2b^2\) = 21-4ab then \(a^2\) = ?

A. (16-4b)/\(b^3\)

B. 28/\(b^3\)

C. 16/(\(b^2\)+7b)

D. 49/\(b^2\)

E. 9/\(b^2\)


\((ab)^2 +4ab-21=0\), Let ab=x
\((x)^2 +4x-21=0\)
(x+7)(x-3)=0
x=-7 or x=3 since x=ab , hence
ab=-7 or ab=3 since a and b,both are negative hence
ab=3 or \(a=\frac{3}{b}\)
\(a^2=\frac{9}{b^2}\)
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If a and b are negative and a   [#permalink] 21 Apr 2019, 12:15
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