Question

What value(s) of b will cause 27x2 + bx + 3 = 0 to have one real solution? Select all that apply.

A. b = −18

B. b = −9

C. b = 9

D. b = 18

Answer 1

Answer:

A & D

Step-by-step explanation:

Just plug in the possible answer choices into the equation to see if it makes sense. You'll find out that -18 and 18 work, while -9 and 9 doesn't.

Answer 2

Quadratic equations can have zero, one or two real solutions

The equation is given as:

$27x^2 + bx + 3 = 0$

For a quadratic equation to have one real solution, the following must be true:

$b^2 = 4ac$

Where:

$a = 27$

$b = b$

$c = 3$

So, we have:

$b^2 = 4ac$

$b^2 = 4 \times 27 \times 3$

$b^2 = 324$

Take square roots of both sides

$b = \sqrt{324$

$b = \±18$

Hence, the values of b are (a) b = -18 and (d) b = 18

Read more about solutions of quadratic equations at:

https://brainly.com/question/3290880