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If ab = 20 – a^2b^2, where both a and b are negative, then

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by leonswati » Thu Sep 15, 2011 12:03 pm
If ab = 20 - a^2b^2, where both a and b are negative, then a^2 =

25/b^2

20-b/b^2

20/b^2

20/b^2+b

16/b^2

I am going mad with this one... Can someone plz solve it....





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by knight247 » Thu Sep 15, 2011 1:08 pm
Hi Swati. Don't go mad yet.lol. Its a very simple one. Read Carefully.

ab=20-a²b²........(1)
a²b²=20-ab
a²=(20-ab)/b²

Let ab=x so a²b²=x² putting these values in our 1st equation we get

x=20-x²
x²+x-20=0
x²+5x-4x-20=0
x(x+5)-4(x+5)=0
(x-4)(x+5)=0
x=4 or x=-5
aka
ab=4 or ab=-5
Now, since we are told that a and b are both negative, their product is going to be +ve so ab=-5 is invalid and therefore discarded. We put the value of ab=4 in the highlighted equation giving

a²=(20-4)/b²=16/b². Hence E
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by leonswati » Thu Sep 15, 2011 7:36 pm
Never thought it this way... Great ya.... thanks....
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by Scott@TargetTestPrep » Sun Jul 28, 2019 9:12 am
leonswati wrote:If ab = 20 - a^2b^2, where both a and b are negative, then a^2 =

25/b^2

20-b/b^2

20/b^2

20/b^2+b

16/b^2
Letting x = ab, we have:

x = 20 - x^2

x^2 + x - 20 = 0

(x - 4)(x + 5) = 0

x = 4 or x = -5

Since both a and b are negative, x = ab will be positive. So x = 4 or ab = 4, and thus a^2*b^2 = 16 and a^2 = 16/b^2.

Answer: E

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