Values such that the system of equations has only one solution

How many different values of a for which the system of equations

$\displaystyle y^2 + y + x^2 = 2$ and $\displaystyle x + y = a$ has only one solution?

Re: Values such that the system of equations has only one solution

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**shiny718** How many different values of a for which the system of equations

$\displaystyle y^2 + y + x^2 = 2$ and $\displaystyle x + y = a$ has only one solution?

From the second equation, we have $\displaystyle \displaystyle \begin{align*} x = a - y \end{align*}$. Substituting into the first equation gives

$\displaystyle \displaystyle \begin{align*} y^2 + y + \left(a - y\right)^2 &= 2 \\ y^2 + y + a^2 - 2a\,y + y^2 &= 2 \\ 2y^2 + \left(1 - 2a\right)y + a^2 - 2 &= 0 \end{align*}$

For there to only be one solution, the discriminant must be 0, so

$\displaystyle \displaystyle \begin{align*} \left(1 - 2a\right)^2 - 4(2)\left(a^2 - 2\right) &= 0 \\ 1 - 4a + 4a^2 - 8a^2 + 16 &= 0 \\ -4a^2 - 4a + 17 &= 0 \end{align*}$

Checking the discriminant of this quadratic tells us how many values of a satisfy this equation, and therefore tell us how many values of a will give one solution to the original set of equations...

$\displaystyle \displaystyle \begin{align*} (-4)^2 - 4(-4)(17) &= 16 + 272 \\ &= 288 \\ &> 0 \end{align*}$

Since this is positive, there are two values of a which will satisfy the set of equations.

Re: Values such that the system of equations has only one solution

Another method. The first equation is that of a circle, which you can verify by representing it in the form (y - y₀)² + x² = r². How many lines with slope -1 touch this circle?

Re: Values such that the system of equations has only one solution

Thanks a lot! :) Thank you for making the solution really clear and simple. :)