In the system of quadratic equations below, c is an integer constant: 2x^2 - 11x + c = 0 x… — SAT

In the system of quadratic equations below, $c$ is an integer constant:

2x^2 - 11x + c = 0

x^2 - 15x + 4c = 0

If the first equation has two distinct real solutions and the second equation has no real solutions, what is the value of $c$ ?

Answer: 15

For the first equation

2x^2 - 11x + c = 0

to have two distinct real solutions, its discriminant must be positive:

\Delta_1 = (-11)^2 - 4(2)(c) = 121 - 8c > 0

, which gives

c < 15.125

. For the second equation

x^2 - 15x + 4c = 0

to have no real solutions, its discriminant must be negative:

\Delta_2 = (-15)^2 - 4(1)(4c) = 225 - 16c < 0

, which gives

c > 14.0625

. The only integer

c

that satisfies the inequality

14.0625 < c < 15.125

15

Set up the inequality for the first equation's discriminant to have two distinct real solutions.

c < 15.125

A quadratic equation

ax^2 + bx + c = 0

has two distinct real solutions if its discriminant,

b^2 - 4ac

, is greater than zero. For

2x^2 - 11x + c = 0

, this means

(-11)^2 - 4(2)(c) > 0

, simplifying to

121 - 8c > 0

, or

c < 15.125

Set up the inequality for the second equation's discriminant to have no real solutions.

c > 14.0625

A quadratic equation has no real solutions if its discriminant is less than zero. For

x^2 - 15x + 4c = 0

, the discriminant is

(-15)^2 - 4(1)(4c) < 0

, simplifying to

225 - 16c < 0

, or

c > 14.0625

Find the integer value of

c

that satisfies both inequalities.

c = 15

Combining the results from the previous steps, we get

14.0625 < c < 15.125

. Since

c

is specified to be an integer, the only integer that lies in this open interval is

15

Using the discriminant to determine the number of real solutions of quadratic equations

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