Question

Use the discriminant to determine the number and type of solutions for the following equation.
\[24x^2 + 13 = 0\]

A. Zero real solutions
B. One rational solution
C. Two rational solutions
D. Two irrational solutions

Answer 1

I have no idea what to do. >_< @beenlightened

Answer 2

ok do you know the method of PEMDAS?

Answer 3

Yes.

Answer 4

jk give me one moment to solve and explain

Answer 5

For any
\(\large\color{black}{ \displaystyle \color{blue}{a}{\rm x}^2+\color{blue}{c} =0 }\)

\(\large\color{black}{ \displaystyle \color{blue}{a}{\rm x}^2=-\color{blue}{c} }\)

\(\large\color{black}{ \displaystyle {\rm x}^2=\frac{-\color{blue}{c}}{\color{blue}{a}} }\)

and that would be an imaginary solution.

Answer 6

Thank you for your solution, but I had already done that much and I would like to ask, is the answer B.?? @SolomonZelman

Answer 7

-c/a is a negative number correct?

Answer 8

Correct.

Answer 9

can you take a square root of a negative number (and get a real solution with this)?

Answer 10

I don't believe so.

Answer 11

The answer must be A. then.

Answer 12

yes, correct, a square root of a negative number will not be equal to any real number

Answer 13

yes, the answer is A.

Answer 14

Got it. Thank you for your explanation. :)

Answer 15

anytime:)
and by the way the actual solution to this would be:

\(\large\color{black}{ \displaystyle 24{\rm x}^2+13=0 }\)

\(\large\color{black}{ \displaystyle 24{\rm x}^2=-13 }\)

\(\large\color{black}{ \displaystyle {\rm x}^2=-\frac{13}{24} }\)

\(\large\color{black}{ \displaystyle {\rm x}=\pm\sqrt{-\frac{13}{24}} }\)

\(\large\color{black}{ \displaystyle {\rm x}=\pm\sqrt{-1}\times \sqrt{\frac{13}{24}} }\)

\(\large\color{black}{ \displaystyle {\rm x}=\pm i\times \sqrt{\frac{13}{24}} }\)

\(\large\color{black}{ \displaystyle {\rm x}=\pm i \sqrt{\frac{13}{24}} }\)

Answer 16

Thank you for the solution. :)

Answer 17

yes, that is an imaginary (not real) solution.

Answer 18

you are welcome!