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If a and b are negative and a^2b^2 = 21 - 4ab then a^2 = ?
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02 Feb 2022, 06:41
02 Feb 2022, 06:41
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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. \(\frac{9}{b^2}\)
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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Re: If a and b are negative and a^2b^2 = 21 - 4ab then a^2 = ?
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02 Feb 2022, 06:54
02 Feb 2022, 06:54
1
Carcass 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. \(\frac{9}{b^2}\)
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
gmatclubot
Re: If a and b are negative and a^2b^2 = 21 - 4ab then a^2 = ? [#permalink]
02 Feb 2022, 06:54
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