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Let c be a real number, and consider the system of quadratic equations

  y = 6x^2 - 9x + c - 3x^2 + 17x

  y = 5x^2 - 3x - 2x^2 + 10x - 18


For which values of  c does this system have:
(a) Exactly one real solution (x,y)
(b) More than one real solution?
(c) No real solutions?

Solutions to the quadratics are (x,y) pairs.

 Feb 18, 2024
 #1
avatar+1720 
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We can analyze the system of quadratic equations to determine the number of real solutions for different values of c. Here's how:

 

Analyzing the Discriminant:

 

The discriminant of a quadratic equation determines the nature of its roots (solutions). It is denoted by the symbol b2−4ac. In this case, considering the first equation (y = 6x^2 - 9x + c):

 

a = 6

 

b = -9

 

c (variable)

 

The discriminant (d) for the first equation is:

 

d = (-9)^2 - 4 * 6 * c

 

The number of real solutions depends on the value of the discriminant:

 

d > 0: Two real and distinct solutions (roots)

 

d = 0: One repeated real solution (root)

 

d < 0: No real solutions (complex roots)

 

Relating Discriminant to c:

 

We want to find the values of c that correspond to each case.

 

(a) Exactly one real solution:

 

For exactly one real solution (repeated root), the discriminant needs to be zero.

 

Therefore, we need to solve:

 

0 = (-9)^2 - 4 * 6 * c

 

This simplifies to:

 

c = \frac{81}{24} = \dfrac{7}{2}

 

(b) More than one real solution:

 

For more than one real solution (distinct roots), the discriminant needs to be positive.

 

Therefore, we need to solve:

 

0 < (-9)^2 - 4 * 6 * c

 

This simplifies to:

 

c < \dfrac{81}{24} = \dfrac{7}{2}

 

(c) No real solutions:

 

For no real solutions (complex roots), the discriminant needs to be negative.

 

Therefore, we need to solve:

 

0 > (-9)^2 - 4 * 6 * c

 

This simplifies to:

 

c > \dfrac{81}{24} = \dfrac{7}{2}

 

Summary:

 

(a) Exactly one real solution: c = dfrac{7}{2}

 

(b) More than one real solution: c < dfrac{7}{2}

 

(c) No real solutions: c > dfrac{7}{2}

 Jul 5, 2024

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