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04 Dec 2019, 02:07
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For which value of k does the following pair of equations yield a unique solution for x, such that the solution is positive?
\((x - k)^2 - y^2 = 8\)
\((x + 2k)^2 + (y^2 - k^2) = 0\)
(A) \(\frac{-4}{\sqrt{7}}\)
(B) \(\frac{\sqrt{7}}{4}\)
(C) \(\sqrt{2}\)
(D) \(\frac{4}{\sqrt{7}}\)
(E) None of these
Are You Up For the Challenge: 700 Level Questions
\((x - k)^2 - y^2 = 8\)
\((x + 2k)^2 + (y^2 - k^2) = 0\)
(A) \(\frac{-4}{\sqrt{7}}\)
(B) \(\frac{\sqrt{7}}{4}\)
(C) \(\sqrt{2}\)
(D) \(\frac{4}{\sqrt{7}}\)
(E) None of these
Are You Up For the Challenge: 700 Level Questions
04 Dec 2019, 05:40
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\((x - k)^2 - y^2 = 8\)
\(y^2= (x-k)^2-8\)......(1)
\((x + 2k)^2 + (y^2 - k^2) = 0\)
From (1)
\((x + 2k)^2 + [(x-k)^2-8 - k^2] = 0\)
\(x^2 + kx + 2k^2 - 4=0\)
We have a unique solution of x; hence, b^2-4ac=0
k^2 - 4 ( 2k^2 - 4 )= 0
k = \(\frac{+4}{\sqrt{7}}\) or \(\frac{-4}{\sqrt{7}}\)
Also, solution is positive; hence, their sum is also positive. k must be negative
Only possible value of k is \(\frac{-4}{\sqrt{7}}\)
\(y^2= (x-k)^2-8\)......(1)
\((x + 2k)^2 + (y^2 - k^2) = 0\)
From (1)
\((x + 2k)^2 + [(x-k)^2-8 - k^2] = 0\)
\(x^2 + kx + 2k^2 - 4=0\)
We have a unique solution of x; hence, b^2-4ac=0
k^2 - 4 ( 2k^2 - 4 )= 0
k = \(\frac{+4}{\sqrt{7}}\) or \(\frac{-4}{\sqrt{7}}\)
Also, solution is positive; hence, their sum is also positive. k must be negative
Only possible value of k is \(\frac{-4}{\sqrt{7}}\)
Bunuel
06 Dec 2019, 08:35
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Bunuel
x = unique positive solution, so \(b^2-4ac=0\) and b<0;
\((x - k)^2 - y^2 = 8\)
\((x + 2k)^2 + (y^2 - k^2) = 0\)
\([x^2+k^2-2xk-y^2=8]+[(x + 2k)^2 + y^2 - k^2 = 0]\)
\([x^2-2xk]+[(x + 2k)^2]=8…x^2-2xk+x^2+4k^2+4xk=8\)
\(2x^2+2xk+4k^2=8…x^2+xk+2k^2=4…x^2+xk+(2k^2-4)=0\)
\(b^2-4ac=0…(xk)^2-4(x^2)(2k^2-4)=0…(xk)^2=4(x^2)(2k^2-4)\)
\(k^2=4(2k^2-4)…8k^2-16-k^2=0…7k^2=16…k^2=16/7\)
\(|k|=[4/√7,-4/√7]…[x^2+xk+(2k^2-4)=0]…[xk]=[b<0]…k=-4/√7\)
Ans (A)
