Hello, we need to solve this system, c being a real number.
[tex]\begin{cases}y &= 3x^2-5x\\y &= 2x^2-x-c\end{cases}[/tex]
y=y, right? So, it comes.
[tex]3x^2-5x=2x^2-x-c\\\\3x^2-2x^2-5x+x+c=0\\\\\boxed{x^2-4x+c=0}[/tex]
We can compute the discriminant.
[tex]\Delta=b^2-4ac=4^2-4c=4(4-c)[/tex]
If the discriminant is 0, there is 1 solution.
It means for [tex]4(4-c)=0 <=> 4-c=0 <=> \boxed{c=4}[/tex]
And the solution is
[tex]x_2=x_1=\dfrac{4}{2}=2[/tex]
If the discriminant is > 0, there are 2 real solutions.
It means 4(4-c) > 0 <=> 4-c > 0 <=> [tex]\boxed{c<4}[/tex]
And the solution are
[tex]x_1=\dfrac{4-\sqrt{4(4-c)}}{2}=\dfrac{4-2\sqrt{4-c}}{2}=2-\sqrt{4-c}\\\\x_2=2+\sqrt{4-c}[/tex]
If the discriminant is < 0, there are no real solutions.
It means 4(4-c) < 0 <=> 4-c < 0 <=> [tex]\boxed{c>4}[/tex]
There are no real solutions and the complex solutions are
[tex]x_1=\dfrac{4-\sqrt{4(4-c)}}{2}=\dfrac{4-2\sqrt{i^2(c-4)}}{2}=2-\sqrt{c-4}\cdot i\\\\x_2=2+\sqrt{c-4}\cdot i[/tex]
Thank you.
