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Manager  S
Joined: 18 Jul 2019
Posts: 54
The following pair of equations have a unique solution for x, which is  [#permalink]

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3 00:00

Difficulty:   55% (hard)

Question Stats: 46% (02:55) correct 54% (03:29) wrong based on 13 sessions

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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)$$\sqrt{10}/9$$

d)$$-\sqrt{10}/9$$

e)$$9/\sqrt{10}$$
Math Expert V
Joined: 02 Aug 2009
Posts: 8289
Re: The following pair of equations have a unique solution for x, which is  [#permalink]

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CaptainLevi wrote:
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)$$9/\sqrt{10}$$

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
_________________ Re: The following pair of equations have a unique solution for x, which is   [#permalink] 25 Nov 2019, 19:45
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# The following pair of equations have a unique solution for x, which is  