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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 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.

Respuesta :

wegnerkolmp2741o

Answer:

see below

Step-by-step explanation:

y = 3x^2 - 5x,

y = 2x^2 - x - c,

Set the two equations equal

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

Subtract 2x^2 from each side

x^2 -5x = -x-c

Add x to each side

x^2 -4x = -c

Add c to each side

x^2 -4x +c = 0

We can use the discriminant

b^2 -4ac  to determine the value for c

a = 1  b = -4  c = c

b^2 - 4ac = 0 means one solution

(-4)^2 - 4(1)c =0

16 -4c = 0

16 = 4c

c=4  is where the is one solution

x^2 -4x-4=0

Factor

(x-2)^2 =0

x=2

y = 3(2)^2 -5(2) = 12 -10 = 2

(2,2) is the solution

When will it have more than one solution

b^2-4ac> 0

(-4)^2 - 4(1)c>0

16 - 4c>0

-4c > -16

Divide each side by -4, remembering to flip the inequality

c< 4

x = -b ± sqrt( b^2-4ac)

     ---------------------------

        2a

y = 3x^2 - 5x

When will it have no solution

b^2-4ac< 0

(-4)^2 - 4(1)c<0

16 -4c<0

-4c<-16

Divide by -4 remembering to flip the inequality

c > 4

no solutions

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