Question

Fan/medal will be rewarded

How do you find the solution of an equation that has "no real solutions" from a negative discriminant.

Answer 1

u can use the quadratic formula, but since the discriminant is negative u can factor the sign inside the square root
\[ \large x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}=\frac{-b\pm\sqrt{(-1)(4ac-b^2)}}{2a} \]
\[ \large =\frac{-b\pm\sqrt{4ac-b^2}\sqrt{-1}}{2a} \]

Answer 2

u just but i insted of sqart -1,and solve like if it was reall.
any way do have a specific prob ??

Answer 3

now the number inside the square root is positive and u get two complex solutions

Answer 4

the problem is -x\[^{2}\] +3x-6=0 @ikram002p

Answer 5

use the quadratic formula

Answer 6

I'm still confused

Answer 7

ok u have
\[ \large -x^2+3x-6=0 \]
right?

Answer 8

yes

Answer 9

so a=-1, b=3, and c=-6
right?

Answer 10

yes

Answer 11

so
\[ \large x=\frac{-3\pm\sqrt{3^2-4(-1)(-6)}}{2(-1)} \]
ok?

Answer 12

Okay

Answer 13

so
\[ \large x=\frac{-3\pm\sqrt{9-24}}{-2}=\frac{-3\pm\sqrt{-15}}{-2} \]
right?

Answer 14

Okay, I'm with you so far.

Answer 15

now
\[ \large x=\frac{-3\pm\sqrt{15(-1)}}{-2}=\frac{-3\pm\sqrt{15}\sqrt{-1}}{-2}
=\frac{-3\pm i\sqrt{15}}{-2} \]
where \(i=\sqrt{-1}\).

Answer 16

Okay, still with you.

Answer 17

that is it. 15 is not a perfect square.

Answer 18

That's it?

Answer 19

yes

Answer 20

Amanda hope u the best :)
helder good job lol.

Answer 21

did u understand? @AmandaMichelle

Answer 22

Thanks @ikram002p and yes it all makes sense now thank you :) @helder_edwin

Answer 23

u r welcome