x^2+y^2=2a+4

x+y=a+3

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

please someone help. thanks

(sorry for bad english)

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- Jun 13th 2012, 03:07 PMTeloParametric 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

please someone help. thanks

(sorry for bad english) - Jun 13th 2012, 03:27 PMPlatoRe: Parametric system of equation problem
- Jun 13th 2012, 03:34 PMTeloRe: 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 PMHallsofIvyRe: 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 $\displaystyle 2x^2- 2ax+ (a^2-2a+ 5)= 0$ Now, an equation like that will have only one root provided it is a**double**root. That is, if it is of the form $\displaystyle 2(x- x_0)^2= 0$. - Jun 13th 2012, 11:22 PMrichard1234Re: 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.

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

Solve the quadratic. - Jun 14th 2012, 01:53 AMTeloRe: Parametric system of equation problem
thank you very much! I've got a=-1

- Jun 14th 2012, 07:35 AMrichard1234Re: 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 AMReckonerRe: Parametric system of equation problem
-1 should be correct. I get a single solution for the resulting system.

My work disagrees slightly with HallsofIvy, however:

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

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

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

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

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

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

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

$\displaystyle \Rightarrow a=-1.$

Substituting this back into the original equations gives the system

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

So

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

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

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

$\displaystyle \Rightarrow x=1$

$\displaystyle \Rightarrow y=1$

A single solution, as desired.