+0  
 
0
251
1
avatar

Consider the quadratic equations \begin{align*} y &=3x^2 - 5x, \\ y &= 2x^2 - x - c \end{align*} where $c$ is a real constant. (a) For what value(s) of $c$ will the system have exactly one solution $(a, b)?$ (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 nonreal solutions?

Guest Mar 8, 2017
Sort: 

1+0 Answers

 #1
avatar+79819 
0

y = 3x^2 - 5x

y = 2x^2 - x - c

 

Set these  =

 

3x^2  - 5x  =  2x^2 - x - c     simplify

 

x^2 - 4x + c  = 0

 

Using the discriminant.....the system will have one real solution whenever

 

4^2  - 4c = 0

4 - c  = 0

4 = c

 

And the system will have more than one real solution whenever

 

4^2 - 4c  > 0

4 - c > 0

4 > c  →    c < 4

 

And the system will have no real solutions whenever

 

c > 4

 

 

 

cool cool cool

CPhill  Mar 8, 2017

6 Online Users

We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details