In the equation x^2 - bx + 16 = 0, b is a positive integer. If the equation has no real so… — SAT

In the equation $x^2 - bx + 16 = 0$ , $b$ is a positive integer. If the equation has no real solutions, what is the greatest possible value of $b$ ?

Answer: 7

The greatest possible value of

b

7

For the quadratic equation

x^2 - bx + 16 = 0

to have no real solutions, its discriminant must be less than

0

. The discriminant is

(-b)^2 - 4(1)(16) = b^2 - 64

. Solving the inequality

b^2 - 64 < 0

gives

b^2 < 64

. Since

b

is a positive integer, taking the square root of both sides gives

b < 8

. The positive integers less than

8

are

1, 2, 3, 4, 5, 6,

and

7

. The greatest of these values is

7

Set up the inequality for the discriminant to be less than zero.

(-b)^2 - 4(1)(16) < 0

A quadratic equation of the form

ax^2 + bx + c = 0

has no real solutions if and only if its discriminant,

D = b^2 - 4ac

, is negative.

Simplify the quadratic inequality.

b^2 < 64

Squaring

-b

yields

b^2

and calculating

4(1)(16)

yields

64

Solve for the greatest positive integer value of

b

b < 8

, so the greatest positive integer is

7

Since

b

is a positive integer, the values satisfying

b^2 < 64

are

1, 2, 3, 4, 5, 6,

and

7

. The largest of these is

7

Quadratic Discriminant and Number of Solutions

Question