y =6x^2 - 9x + c, y = 5x^2 - 3x

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

For which values of does this system have:

(a) Exactly one real solution

(b) More than one real solution?

(c) No real solutions?

gaberiellz Apr 15, 2024

#1**+1 **

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}

Boseo Apr 15, 2024