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

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Bunuel
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If a and b are negative and a^2b^2 = 21 - 4ab then a^2 = ? [#permalink] New post   26 Nov 2021, 04:15
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If a and b are negative and \(a^2b^2 = 21 - 4ab\) then \(a^2 = ?\)

A. \(\frac{16-4b}{b^3}\)

B. \(\frac{28}{b^3}\)

C. \(\frac{16}{b^2+7b}\)

D. \(\frac{49}{b^2}\)

E. \(\frac{9}{b^2}\)
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E
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Re: If a and b are negative and a^2b^2 = 21 - 4ab then a^2 = ? [#permalink] New post   26 Nov 2021, 05:04
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Bunuel wrote:
If a and b are negative and \(a^2b^2 = 21 - 4ab\) then \(a^2 = ?\)

A. \(\frac{16-4b}{b^3}\)

B. \(\frac{28}{b^3}\)

C. \(\frac{16}{b^2+7b}\)

D. \(\frac{49}{b^2}\)

E. \(\frac{9}{b^2}\)


Given:
a and b are negative.
&
a^2b^2 = 21 -4ab
We need to find the value of a^2

Solving question stem:
a^2b^2 = 21 -4ab can be recalibrated as a^2b^2+4ab-21=0
a^2b^2 + 7ab -3ab -21 =0
ab(ab+7)-3(ab+7)=0
So, ab = 3 or -7
It cannot be -7 as a and b are -ve and product of two -ve terms is always +ve.
So ab=3

(ab)^2 = 9
a^2b^2 = 9
a^2 = 9/b^2 E is the ans
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Re: If a and b are negative and a^2b^2 = 21 - 4ab then a^2 = ? [#permalink] New post   26 Nov 2021, 05:36
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Bunuel wrote:
If a and b are negative and \(a^2b^2 = 21 - 4ab\) then \(a^2 = ?\)

A. \(\frac{16-4b}{b^3}\)

B. \(\frac{28}{b^3}\)

C. \(\frac{16}{b^2+7b}\)

D. \(\frac{49}{b^2}\)

E. \(\frac{9}{b^2}\)


Solution:

Let us assume \(ab=x\).

So, the equation \(a^2b^2 = 21 - 4ab\) will become \(x^2=21-4x\) or \(x^2+4x-21=0\).

We can solve \(x^2+4x-21=0\) like we solve any other quadratic equation.

\(⇒ x^2+7x-3x-21=0\)

\(⇒ x(x+7)-3(x+7)=0\)

\(⇒ (x+7)(x-3)=0\)

So the value of x can be either \(-7\) or \(3\).

We need to keep in mind that \(x=ab\) and a and b are negative numbers. This means \(ab\) will be positive.

Thus, \(ab=-7\) is discarded.

We have \(ab=3\)

\(⇒ a=\frac{3}{b}\)

\(⇒ a^2 = (\frac{3}{b})^2\)

\(⇒ a^2 = \frac{9}{b^2}\)

Hence the right answer is Option E.
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If a and b are negative and a^2b^2 = 21 - 4ab then a^2 = ? [#permalink] New post   Updated on: 26 Nov 2021, 08:07
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Bunuel wrote:
If a and b are negative and \(a^2b^2 = 21 - 4ab\) then \(a^2 = ?\)

A. \(\frac{16-4b}{b^3}\)

B. \(\frac{28}{b^3}\)

C. \(\frac{16}{b^2+7b}\)

D. \(\frac{49}{b^2}\)

E. \(\frac{9}{b^2}\)


Given: \(a^2b^2 = 21 - 4ab\)
Since this looks like a quadratic equation, let's set it equal to zero: \(a^2b^2 + 4ab - 21 = 0\)
Factor to get: \((ab +7)(ab -3) = 0\)

So, EITHER \(ab = -7\) OR \(ab =3\)
If \(\)a and \(b\) are negative, then the product \(ab\) must be positive, which means it must be the case that \(ab =3\)

If \(ab =4\), then \(a = \frac{3}{b}\)
Square both sides to get: \(a^2 = (\frac{3}{b})^2 = \frac{9}{b^2}\)

Answer: E

Originally posted by BrentGMATPrepNow on 26 Nov 2021, 07:05.
Last edited by BrentGMATPrepNow on 26 Nov 2021, 08:07, edited 1 time in total.
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Re: If a and b are negative and a^2b^2 = 21 - 4ab then a^2 = ? [#permalink] New post   26 Nov 2021, 08:00
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BrentGMATPrepNow wrote:
Bunuel wrote:
If a and b are negative and \(a^2b^2 = 21 - 4ab\) then \(a^2 = ?\)

A. \(\frac{16-4b}{b^3}\)

B. \(\frac{28}{b^3}\)

C. \(\frac{16}{b^2+7b}\)

D. \(\frac{49}{b^2}\)

E. \(\frac{9}{b^2}\)


Given: \(a^2b^2 = 21 - 4ab\)
Since this looks like a quadratic equation, let's set it equal to zero: \(a^2b^2 + 4ab - 21 = 0\)
Factor to get: \((ab +7)(ab -4) = 0\)

So, EITHER \(ab = -7\) OR \(ab =4\)
If \(\)a and \(b\) are negative, then the product \(ab\) must be positive, which means it must be the case that \(ab =4\)

If \(ab =4\), then \(a = \frac{4}{b}\)
Square both sides to get: \(a^2 = (\frac{4}{b})^2 = \frac{16}{b^2}\)

Answer: C


Hi Brent, my intention is not to nose-pick but still please check the factors you have made for the factorisation equation. 7*4 =28 & 7*3 21 and we have 21 in the equation.(It happens to me all the time so I understand)
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Re: If a and b are negative and a^2b^2 = 21 - 4ab then a^2 = ? [#permalink] New post   26 Nov 2021, 08:12
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ElninoEffect wrote:
Hi Brent, my intention is not to nose-pick but still please check the factors you have made for the factorisation equation. 7*4 =28 & 7*3 21 and we have 21 in the equation.(It happens to me all the time so I understand)


Silly mistake. Good catch!
I've edited my response accordingly.

Kudos for you!!
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Re: If a and b are negative and a^2b^2 = 21 - 4ab then a^2 = ? [#permalink] New post   26 Nov 2021, 09:09
BrentGMATPrepNow wrote:
ElninoEffect wrote:
Hi Brent, my intention is not to nose-pick but still please check the factors you have made for the factorisation equation. 7*4 =28 & 7*3 21 and we have 21 in the equation.(It happens to me all the time so I understand)


Silly mistake. Good catch!
I've edited my response accordingly.

Kudos for you!!


We learn from each other. :)
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Re: If a and b are negative and a^2b^2 = 21 - 4ab then a^2 = ? [#permalink] New post   28 Mar 2024, 06:13
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