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.
+14915 The system of quadratic equations y = 3x^2 - 5x, y = 2x^2 - x - c
c is a real number.
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\( y = 3x^2 - 5x = 2x^2 - x - c\ |\ Equate\ equation\ to\ zero.\ |\color{blue} \ -(2x^2-x-c)\\ x^2-4x+c =0\ |\ p-q\ formula:\color{blue}x=-\frac{p}{2}\pm\sqrt{(\frac{p}{2})^2-q}\\ x=\frac{4}{2}\pm\sqrt{(\frac{4}{2})^2-c}\\\)
(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:
\((\frac{4}{2})^2-c<0\\ \color{blue}c>4 \)
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