How to solve $x^2-(a+1)x+2a-2=0$ for every parameter $a$ from $\mathbb R$

Question

I need to solve this equation. It says which number $x$ is the solution to this equation for every parameter $a$? I know how to solve parametric equation, however, i don't know which condition i should use here? I tried to find a single number x so i solved $x_1=x_2$ but the solution is not every $a$ from $\mathbb R$.

Answer 1

Discriminant:

$$\Delta=(a+1)^2-4(2a-2)=a^2-6a+9=(a-3)^2$$

and your solution is, with the quadratic formula

$$x_{1,2}=\frac{(a+1)\pm(a-3)}2$$

Finish now the argument.

Answer 2

using the formula for a quadratic equation we get $$x_{1,2}=\frac{a+1}{2}\pm\sqrt{(\frac{a+1}{2})^2-2a+2}$$ simplifying this we get $$x_1=2$$ or $$x_2=a-1$$

Answer 3

If you solve the equation you get that $x_1=a-1$ and $x_2=2$. Thus, independently from the value of $a$, $2$ is the solution of this equation.

Answer 4

Rewriting $-(a+1)x$ using the factors of $2(a-1)$:

$$\begin{eqnarray} x^2-(a+1)x+2(a-1)&=0 \\\Rightarrow x^2-2x-(a-1)x+2(a-1)&=0 \\ \Rightarrow x(x-2)-(a-1)(x-2)&=0 \\ \Rightarrow (x-2)(x-(a-1))&=0. \end{eqnarray}$$

Using no-zero-divisors:

$$\begin{eqnarray} \Rightarrow x-2=0&\text{ or }&x-(a-1)=0 \\ \Rightarrow x=2&\text{ or }&x=a-1, \end{eqnarray}$$

so that $x=2$ is the answer to the question that you asked.