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Joined: 20 Feb 2017
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If x and y are positive and x^2 * y^2 = 18 – 3xy, then x^2 =?
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10 May 2021, 21:43
10 May 2021, 21:43
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If x and y are positive and \(x^2 * y^2 = 18 – 3xy,\) then \(x^2 =\)?
A. \(\frac{(18-3y)}{y^3}\)
B. \(\frac{18}{y^2}\)
C. \(\frac{18}{(y^2+3y)}\)
D. \(\frac{9}{y^2}\)
E. \(\frac{36}{y^2}\)
A. \(\frac{(18-3y)}{y^3}\)
B. \(\frac{18}{y^2}\)
C. \(\frac{18}{(y^2+3y)}\)
D. \(\frac{9}{y^2}\)
E. \(\frac{36}{y^2}\)
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Re: If x and y are positive and x^2 * y^2 = 18 – 3xy, then x^2 =?
[#permalink]
16 May 2021, 06:28
16 May 2021, 06:28
2
\(x^2 * y^2 = 18 – 3xy,\)
Lets rewrite the equation as \((xy)^2 + 3xy - 18= 0\)
Now lets take xy as T we get
\(T^2 + 3T - 18= 0\)
\(T^2 + 6T - 3T - 18= 0\)
\(T(T + 6) - 3(T + 6)= 0\)
\((T + 6)(T - 3)= 0\)
=> T = 3, -6
=> xy = 3 or xy =-6
=> x = \(\frac{3}{y}\)
Squaring both sides we get
\(x^2\) = \(\frac{9}{y^2}\)
So, answer will be D
Hope it helps!
Lets rewrite the equation as \((xy)^2 + 3xy - 18= 0\)
Now lets take xy as T we get
\(T^2 + 3T - 18= 0\)
\(T^2 + 6T - 3T - 18= 0\)
\(T(T + 6) - 3(T + 6)= 0\)
\((T + 6)(T - 3)= 0\)
=> T = 3, -6
=> xy = 3 or xy =-6
=> x = \(\frac{3}{y}\)
Squaring both sides we get
\(x^2\) = \(\frac{9}{y^2}\)
So, answer will be D
Hope it helps!
gmatclubot
Re: If x and y are positive and x^2 * y^2 = 18 – 3xy, then x^2 =? [#permalink]
16 May 2021, 06:28
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