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)
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 $\displaystyle 2(x- x_0)^2= 0$.
-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.