Question

Check:

Solve 3x2 + 4x + 2 = 0. Round solutions to the nearest hundredth.

x ≈ −0.59 and x ≈ −3.41

x ≈ 0.20 and x ≈ 1.14

x ≈ −0.20 and x ≈ −1.14

No real solutions

Answer 1

\[x = \frac{ -4 \pm \sqrt{4^{2} - 4(3)(2)} }{ 2(3) }\]

Answer 2

check the discriminant value (D), if D < 0 then no real solutions
with D = b^2 - 4ac

Answer 3

Use the discriminant to determine is you can solve it:
\[\sqrt{4^2-4(3)(2)}\]

\[\sqrt{16-24}\]

Answer 4

\[x = \frac{ -4 \pm \sqrt{16 - 24} }{ 2(3) }\]

Answer 5

\[x = \frac{ -4 \pm \sqrt{-8} }{ 2(3) }\]

Answer 6

\[x = \frac{ -4 \pm 2.82 }{ 6 }\]

Answer 7

Right so far?

Answer 8

\[x = \frac{ -4 + 2.82 }{ 6 }\]

\[x = \frac{ -4 - 2.82 }{ 6 }\]

Answer 9

sqrt(-8), it is a complex number

Answer 10

Yes but you would shorten it and do \[\approx\] right?

Answer 11

When you have a negative in a radical you get imaginary solutions.

Answer 12

\[x \approx \frac{ -1.18 }{ 6 }\]

\[x \approx \frac{ -6.82 }{ 6 }\]

Answer 13

\[x \approx 0.196\]

\[x \approx -1.136\]

Answer 14

So this would be "No real solution" right?

Answer 15

I am confused <_<

Answer 16

Yes, there are no real solutions.. and oh hey JA didn't realize it was you! :D

Answer 17

lol hey :)

Answer 18

Wait but how could you know before solving it?

Answer 19

By checking the discriminant

Answer 20

? I am very bad with mathematical terms :/

Answer 21

The \[\sqrt{b^2-4ac}\]
:P

Answer 22

Yes yes but how would you see which part tells you?

Answer 23

If it's positive then it'll have real solutions
If it's negative it has imaginary solutions

Answer 24

OHHHHHH I see now

Answer 25

Blind as a bat <_<

Answer 26

Naw :P

Answer 27

lol k thanks a heap :)

Answer 28

Yup!