BRAINLIEST AND MANY POINTS TO BEST ANSWER (has to have complete sentences, thorough explanations, and answer all parts of the question) Thank you so much!! 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.

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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step-by-step explanation:

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