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The following pair of equations have a unique solution for x, which is negative. For what value of p does this happen?
\((3p – x)^2 – 1 + y^2 = 0\)
\(9x^2 – y^2 = 0\)
(A) 0
(B) 3
(C) \(\frac{\sqrt{10}}{9}\)
(D) \(-\frac{\sqrt{10}}{9}\)
(E) \(\frac{9}{\sqrt{10}}\)
\((3p – x)^2 – 1 + y^2 = 0\)
\(9x^2 – y^2 = 0\)
(A) 0
(B) 3
(C) \(\frac{\sqrt{10}}{9}\)
(D) \(-\frac{\sqrt{10}}{9}\)
(E) \(\frac{9}{\sqrt{10}}\)
25 Nov 2019, 19:45
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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?
\((3p – x)^2 – 1 + y^2 = 0\)
\(9x^2 – y^2 = 0\)
a) 0
b) 3
c)\(\sqrt{10}/9\)
d)\(-\sqrt{10}/9\)
e)\(9/\sqrt{10}\)
\((3p – x)^2 – 1 + y^2 = 0\)
\(9x^2 – y^2 = 0\)
a) 0
b) 3
c)\(\sqrt{10}/9\)
d)\(-\sqrt{10}/9\)
e)\(9/\sqrt{10}\)
Add both the equations to get rid of y..
\((3p – x)^2 – 1 + y^2 +9x^2 – y^2 = 0....9p^2-6px+x^2+9x^2-1=0.....10x^2-6px+(9p^2-1)=0\)
Now for this to have a unique negative solution, it should be able to be written in the form \((x+a)^2=x^2+2ax+a^2\)
So the value of p has to be NEGATIVE to make -6px positive...ONLY D possible..
If you have time and you want to solve it further
\(10x^2-6px+(9p^2-1)=x^2-\frac{6px}{10}+\frac{9p^2-1}{10}=x^2+2ax+a^2\)
Thus \(\frac{-6px}{10}=2ax......a=\frac{-3p}{10}\)
Also \(\frac{9p^2-1}{10}=a^2=(\frac{3p}{10})^2=\frac{9p^2}{100}.......90p^2-10=9p^2.......81p^2=10....9p=\sqrt{10}..or -\sqrt{10}....p=\frac{-\sqrt{10}}{9}\) as p is negative
D
