# Parametric system of equation problem

• Jun 13th 2012, 03:07 PM
Telo
Parametric system of equation problem
x^2+y^2=2a+4
x+y=a+3

find the meaning for "a" when system of equation has only 1 root

• Jun 13th 2012, 03:27 PM
Plato
Re: Parametric system of equation problem
Quote:

Originally Posted by Telo
x^2+y^2=2a+4
x+y=a+3
find the meaning for "a" when system of equation has only 1 root

Solve $x^2+(a+3-x)^2=2a+4$ for $x$.
• Jun 13th 2012, 03:34 PM
Telo
Re: Parametric system of equation problem
thanks for answer but i didnt get it, i have this: 2x^2-2ax+a^2+9=2a+4 what do i do next? how do i find out what meaning must "a" have? for 1 root of x
• Jun 13th 2012, 03:51 PM
HallsofIvy
Re: Parametric system of equation problem
Do you not know how to solve a quadratic equation?

But more to the point, that equation can be written $2x^2- 2ax+ (a^2-2a+ 5)= 0$ Now, an equation like that will have only one root provided it is a double root. That is, if it is of the form $2(x- x_0)^2= 0$.
• Jun 13th 2012, 11:22 PM
richard1234
Re: Parametric system of equation problem
Expanding on HallsofIvy's post, you have a double root if and only if the discriminant is zero, i.e.

$(-2a)^2 - 4(2)(a^2 - 2a + 5) = 0$

• Jun 14th 2012, 01:53 AM
Telo
Re: Parametric system of equation problem
thank you very much! I've got a=-1
• Jun 14th 2012, 07:35 AM
richard1234
Re: Parametric system of equation problem
Hmm I'm not sure if -1 is correct...according to the quadratic I posted, it has two non-real roots for a.
• Jun 14th 2012, 08:29 AM
Reckoner
Re: Parametric system of equation problem
Quote:

Originally Posted by richard1234
Hmm I'm not sure if -1 is correct...according to the quadratic I posted, it has two non-real roots for a.

-1 should be correct. I get a single solution for the resulting system.

My work disagrees slightly with HallsofIvy, however:

\begin{align*}x^2+y^2&=2a+4\\x+y&=a+3\end{align*}

$\Rightarrow x^2 + \left(a-x+3\right)^2=2a+4$

$\Rightarrow2x^2-2ax-6x+a^2+6a+9=2a+4$

$\Rightarrow2x^2-2(a+3)x+\left(a^2+4a+5\right)=0$

For this quadratic to have a single root, the discriminant $b^2-4ac$ must be 0:

$4(a+3)^2-8\left(a^2+4a+5\right)=0$

$\Rightarrow a^2+2a+1=0$

$\Rightarrow a=-1.$

Substituting this back into the original equations gives the system

\begin{align*}x^2+y^2&=2\\x+y&=2\end{align*}

So

$x^2 + (2-x)^2=2$

$\Rightarrow x^2-2x+1=0$

$\Rightarrow(x-1)^2=0$

$\Rightarrow x=1$

$\Rightarrow y=1$

A single solution, as desired.