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Consider the system of quadratic equations y = 3x^2 - 5x, 
                                                                      y = 2x^2 - x - c,
where c is a real number.

(a) For what value(s) of c will the system have exactly one real 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.

 

 

please do the most annoying thing of showing your work.
 

 Sep 1, 2022
 #1
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The system of quadratic equations     y = 3x^2 - 5x,      y = 2x^2 - x - c
c is a real number.

 

Hello Guest!

 

\( y = 3x^2 - 5x = 2x^2 - x - c\ |\ Equate\ equation\ to\ zero.\ |\color{blue} \ -(2x^2-x-c)\\ x^2-4x+c =0\ |\ p-q\ formula:\color{blue}x=-\frac{p}{2}\pm\sqrt{(\frac{p}{2})^2-q}\\ x=\frac{4}{2}\pm\sqrt{(\frac{4}{2})^2-c}\\\)

(a)

The system have exactly one real solution: 

\((\frac{4}{2})^2-c=0\\ c=(\frac{4}{2})^2\\ \color{blue}c=4\)

(b)

The system will have more than one real solution:

\((\frac{4}{2})^2-c>0 \)

\(\color{blue}\infty\) < c < 4

(c)

The system will have no real solutions:

\((\frac{4}{2})^2-c<0\\ \color{blue}c>4 \)

laugh  !

 Sep 2, 2022

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