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CaptainLevi
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CaptainLevi

The following pair of equations have a unique solution for x, which is negative. For what value of p does this happen?

\((3p – x)^2 – 1 + y^2 = 0\)

\(9x^2 – y^2 = 0\)

\(y^2 = 9x^2\)

\(9p^2 - 6px + x^2 + 9x^2 - 1 = 0\)
\(10x^2 - 6px + (9p^2-1) = 0\)
\(x^2 - 3px/5 + (9p^2 - 1)/10 = 0\)

Since x is negative, the equation should be of the form (x+a)^2 = 0; where a is positive

\(x^2 - 3px/5 + (9p^2 - 1)/10 = x^2 + 2ax + a^2 \)
-3p/5 = 2a; p = -10a/3 < 0 ; since a>0
a = -3p/10

\((9p^2-1)/10 = a^2 = 9p^2/100\)
\(90p^2 - 10 = 9p^2\)
\(81p^2 = 10\)

\(p = - \frac{\sqrt{10}}{9}\) since p is negative

(A) 0;

(B) 3;

(C) \(\frac{\sqrt{10}}{9}\)

(D) \(-\frac{\sqrt{10}}{9}\)

(E) \(\frac{9}{\sqrt{10}}\)


IMO D
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