Consider the system of quadratic equations \begin{align} y &=3x^2 - 5x, \\ y &= 2x^2 - x - c, \end{align}where $c$ is a real number. (a) For what value(s) of $c$ will the system have exactly one solution $(x,y)?$ (b) For what value(s) of $c$ will the system have more than one real solution? (c) For what value(s) of $c$ will the system have no real solutions? Solutions to the quadratics are $(x,y)$ pairs. Your answers will be in terms of $c,$ but make sure you address both $x$ and $y$ for each part.

Hello, we need to solve this system, c being a real number.

$\begin{cases}y &= 3x^2-5x\\y &= 2x^2-x-c\end{cases}$

y=y, right? So, it comes.

$3x^2-5x=2x^2-x-c\\\\3x^2-2x^2-5x+x+c=0\\\\\boxed{x^2-4x+c=0}$

We can compute the discriminant.

$\Delta=b^2-4ac=4^2-4c=4(4-c)$

If the discriminant is 0, there is 1 solution.

It means for $4(4-c)=0 4-c=0 \boxed{c=4}$

And the solution is

$x_2=x_1=\dfrac{4}{2}=2$

If the discriminant is > 0, there are 2 real solutions.

It means 4(4-c) > 0 <=> 4-c > 0 <=> $\boxed{c$

And the solution are

$x_1=\dfrac{4-\sqrt{4(4-c)}}{2}=\dfrac{4-2\sqrt{4-c}}{2}=2-\sqrt{4-c}\\\\x_2=2+\sqrt{4-c}$

If the discriminant is < 0, there are no real solutions.

It means 4(4-c) < 0 <=> 4-c < 0 <=> $\boxed{c>4}$

There are no real solutions and the complex solutions are

$x_1=\dfrac{4-\sqrt{4(4-c)}}{2}=\dfrac{4-2\sqrt{i^2(c-4)}}{2}=2-\sqrt{c-4}\cdot i\\\\x_2=2+\sqrt{c-4}\cdot i$

Thank you.

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