Question

How would you use the quadratic formula to solve x^2+5x=-2. Using compete sentences.

Answer 1

@SolomonZelman

Answer 2

maybe we can work the prob using math, and you do the sentences part alone?

Answer 3

Okay

Answer 4

when the equation is \(\LARGE\color{black}{ \color{blue}{a} x^2+ \color{magenta}{b}x+ \color{red}{c}=0 }\).


you use the following formula:
\(\normalsize\color{ slate }{\Huge{\bbox[5pt, lightcyan ,border:2px solid black ]{ \LARGE{x=~} \huge{ \frac{-\color{magenta}{b} \pm\sqrt{ \color{magenta}{b} ^2-4 \color{blue}{a} \color{red}{c}}}{2 \color{blue}{a}} }~ }}}\)

(to transfer your equation to the needed \(\large \color{black}{ \color{blue}{a} x^2+ \color{magenta}{b}x+ \color{red}{c}=0 }\) form, you have to add 2 to both sides.)

Answer 5

So it will be x^2+5+2=0

Answer 6

what do you get after adding 2 to both sides?

Answer 7

yes

Answer 8

\(\LARGE\color{black}{ \color{blue}{a} x^2+ \color{magenta}{b}x+ \color{red}{c}=0 }\)

` ... comparing ... `

\(\LARGE\color{black}{ \color{blue}{1} x^2+ \color{magenta}{5}x+ \color{red}{2}=0 }\)

Answer 9

Now, please plug in the a b c for me....

Answer 10

What do you mean by plug in?

Answer 11

I have no clue how to do this. I am just learning it

Answer 12

she meant let a = 1, b = 5, and c = 2
we will have this equation \[x^2+5x+2=0\]

Answer 13

Okay and then what do I do from there?

Answer 14

Well, to solve this faster we need the discriminant \[b^2-4ac\] so let b = 5, a =1, and c =2

Answer 15

who is she?

Answer 16

umm I thought that was you?

Answer 17

well, if you are thinking of me as of a good looking girl, rather than a bad looking girl, then you in a sense did tackle my identity.

Answer 18

anyway plug in b =5, c =2, and a = 1 into the discriminant formula \[b^2-4ac\] @J_slate23

Answer 19

as in replace the b with 5, replace the c with 2 and replace the a with 1

Answer 20

So 5^2=4(1)(2)?

Answer 21

yes but replace the = with -

Answer 22

\[5^2-4(1)(2)\]

Answer 23

So 5^2-4(1)(2)?

Answer 24

mhm so now what's 5^2
and what's 4(1)(2)

Answer 25

\(\normalsize\color{ slate }{\Huge{\bbox[5pt, lightyellow ,border:2px solid black ]{ \LARGE{x=~} \huge{ \frac{-\color{magenta}{5} \pm\sqrt{ \color{magenta}{5} ^2-4 \color{blue}{(1)} \color{red}{(2)}}}{2 \color{blue}{(1)}} }~ }}}\)

and your equation is \(\LARGE\color{black}{ \color{blue}{1} x^2+ \color{magenta}{5}x+ \color{red}{2}=0 }\).

Answer 26

25 and 8

Answer 27

yes so now what's 25-8

Answer 28

17

Answer 29

yes! so for the bottom denominator what is 2(1) ?

Answer 30

2

Answer 31

yes

Answer 32

so now you have \[\frac{5 +\sqrt{17}}{2}, \frac{5-\sqrt{17}}{2}\] I split the signs up because the latex for the quadratic formula is big and nasty

Answer 33

OH MAN I forgot the -!

Answer 34

\[\frac{-5 +\sqrt{17}}{2}, \frac{-5-\sqrt{17}}{2}\]

Answer 35

Is that the answer? If so that was really easy

Answer 36

yeah.. I just split the sign in the middle up. the latex for the entire quadratic equation with + and - together was a pain

Answer 37

and 17 is not a perfect square... so we can't simplify .. not to mention 17 is a prime

Answer 38

So the final answer is -5+-sqrt17/2?

Answer 39

yeah

Answer 40

Thank you!

Answer 41

That is easier then I expected

Answer 42

Can you check my math for another one? @UskiDoll

Answer 43

sure

Answer 44

Solve x^2+4x-12=0 by completing the square.

X^2+4x-12=0
Add 12
X^2+4x=12
(X+2)^2-4=12
(x+2)^2=16
(x+2)= 4 or -4
X= 6 or -6 is my answer

Answer 45

Is that correct @UskiDoll

Answer 46

\[x^2+4x-12=0 \]
\[x^2+4x=12 \]
\[4 \times \frac{1}{2} = \frac{4}{2} = 2\]
\[2^2 = 4\]
\[x^2+4x+4=12+4 \]
\[x^2+4x+4=16 \]
\[(x+2)^2 = 16 \]
square root on both sides
\[\sqrt{(x+2)^2} = \sqrt{16}\]
\[x+2=4, x+2=-4\]
\[x=2, x=-6\]

Answer 47

x+2 = 4 on your part is wrong. subtract 2 from both sides, but -6 is correct

Answer 48

Okay. Thank you

Answer 49

plug it in -b+_ b2-4ac \[\sqrt{b2-4ac}\] / 4a good luck

Answer 50

The problem is already done ^_^