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08 Feb 2021, 05:29
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How to answer this:
x, y > 0
\(6x^2 + 5xy - 4y^2 = 0\)
\(\frac{y}{x} = ?\)
x, y > 0
\(6x^2 + 5xy - 4y^2 = 0\)
\(\frac{y}{x} = ?\)
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08 Feb 2021, 08:46
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iatii
Divide the equation with \(x^2\)
\(\frac{6x^2}{x^2} + \frac{5xy}{x^2} - \frac{4y^2}{x^2} = 0\)
\(6 + \frac{5y}{x} - \frac{4y^2}{x^2} = 0\)
Let \(\frac{y}{x}\) be \(z\)
\(6 + 5z - 4z^2 = 0\)
\(4z^2 - 5z - 6 = 0\)
\(4z^2 - 8z + 3z - 6 = 0\)
\(4z(z - 2) + 3(z - 2) = 0\)
\((z - 2)(4z + 3) = 0\)
z = 2 or -3/4
\(\frac{y}{x}\) = 2 or -3/4
Since x, y > 0, \(\frac{y}{x}\) = 2
08 Feb 2021, 09:26
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Given: x, y > 0
\(6x^2 + 5xy - 4y^2 = 0\)
\(\frac{y}{x} = ?\)
\(6x^2 + 5xy - 4y^2 = 0\)
Let y/x = t; y=xt
6x^2 + 5tx^2 - 4t^2x^2 = 0
6 + 5t - 4t^2 = 0
6 + 8t - 3t - 4t^2 = 0
(2-t)(3+4t) =0
t = 2 or - 3/4
Since x,y>0
t = 2
\(6x^2 + 5xy - 4y^2 = 0\)
\(\frac{y}{x} = ?\)
\(6x^2 + 5xy - 4y^2 = 0\)
Let y/x = t; y=xt
6x^2 + 5tx^2 - 4t^2x^2 = 0
6 + 5t - 4t^2 = 0
6 + 8t - 3t - 4t^2 = 0
(2-t)(3+4t) =0
t = 2 or - 3/4
Since x,y>0
t = 2
