Question

Solve 2x2 + 3x + 5 = 0. Round solutions to the nearest hundredth I got -4.75 and -1.25

Answer 1

need to make sure I solved correctly

Answer 2

\[
2x^2 + 3x + 5 = 0\\
b^2 -4 a c=3^2 -4(2)(5)=9 -40=-31\\
\]
So the two roots are complex.

Answer 3

\[

\begin{array}{c}
x=\frac{1}{4} \left(-3-i
\sqrt{31}\right) \\
x=\frac{1}{4} \left(-3+i
\sqrt{31}\right) \\
\end{array}

\]

Answer 4

So there is no real solution?

Answer 5

No

Answer 6

Did you type your problem correctly?

Answer 7

2x^2+3x+5=0

Answer 8

May be they want you to write the roots as
\[

\begin{array}{c}
x=-0.75-1.39194 i \\
x=-0.75+1.39194 i \\
\end{array}

\]

Answer 9

To the nearest hundredth
\[

\begin{array}{c}
x=-0.75-1.39 i \\
x=-0.75+1.39 i \\
\end{array}

\]

Answer 10

Are you still there?

Answer 11

Yes I'm just confused as to how you got that answer

Answer 12

Use the quadratic formula and your Calculator.

Answer 13

Oh I did +4ac instead of -4ac

Answer 14

A big difference.

Answer 15

I'm getting a -31 and you can't square - numbers

Answer 16

?

Answer 17

\[
\sqrt{-31} = i \sqrt{31}
\]

Answer 18

Where
\[
i^2 = -1
\]

Answer 19

What does that mean?

Answer 20

Do you know how to use the quadratic formula? Do you know what a complex number is?

Answer 21

quadratic formula , yes
complex number, no

Answer 22

For quadratic formula, you need to know complex numbers in case b^2 - 4 ac <0

Answer 23

Okay and what if you do have a complex number?