Question

find the solution to the quadratic equation
x^2 + x = 3x - 5

Answer 1

it's right there in front of you!

did you mean find the solutions to the quadratic equation?

Answer 2

oops , yes the solution! find the solution!

Answer 3

you need to use the quadratic formula on:

\[ax^2 + bx + c = 0\]

\[x_{1,2} = \frac{ -b \pm \sqrt{b^2 - 4ac} }{ 2a }\]

you get 2 x values. one for plus, one for minus on the plusminus sign

Answer 4

This is also pretty easy to solve by completing the square, if you prefer.

Answer 5

these are the possible solution


A. 5±√21
_______
2

B. 3±√34
_______
3
C.-2±√-16
_______
2
D.2±√-16
_______
2
E.-2±√21
_______
4
F.-5±√6
_______
2

Answer 6

someone clarify pleeeeasee

Answer 7

move everything to one side and then plug in values to the quadratic formula

Answer 8

my answer is D

Answer 9

D is correct

Answer 10

luigi!

Answer 11

nice score! :)

Answer 12

Using @whpalmer4's way, let's try completing the square.

\[\large x^2 +x =3x -5\]\[\large x^2 + x - 3x = -5\]\[\large x^2 -2x = -5\]\[
\quad \large c=(\frac{-2}{2})^2=1\]\[\large x^2 - 2x +1 = -5+1\]\[\large (x-1)^2 = -4\]\[\large \sqrt{(x-1)^2}=\sqrt{-4}\]\[\large x-1 = \pm 2i\]\[\large x= 1 \pm2i \implies x- 2i, x+2i\]

Answer 13

I would have stopped at \(x=1\pm2i\), not sure what the rightarrow is supposed to indicate...

Answer 14

you will all surpass me eventually, dan :P

Answer 15

"implies"

Answer 16

no i wont i will wait for you level then i will lvl

Answer 17

okay, but I don't see how it implies what you have to the right of the arrow :-)

Answer 18

@dan815 mine is better. ;) lolol.

Answer 19

69 is that akward phase no one wants to be in

Answer 20

its like being 18 before u can drink

Answer 21

well you're gonna have to wait awhile then :P

Answer 22

what i mean is if you break it down
"x pm 2i" means x=1+ 2i and x= 1-2i lol.

Answer 23

x^2 + x = 3x - 5
first solve so as to equal zero
x^2 -2x +5 = 0
The quadratic takes different forms, in engineering its
0 = x^2 +bx +c = -(b/2)(1 +- sqrt(1 - c(2/b)^2) :b =B/A. c = C/A
therefore
x = 1(1 +- sqrt(1 - 5(1^2)) = 1 +- i4
when complex always come in conjugate form
(x - 1 + i4)(x -1 - i4) = 0

Answer 24

@dan815 watch, Jhanny will even pass me.. then she'll be your competition :P

Answer 25

I don't feel smart enough!

Answer 26

@kenljw but what about that pesky detail that \[(x-1+4i)(x-1-4i) = x^2-2x+17\]

Answer 27

sqrt(1-5(1^2) = 2i, not 4i

Answer 28

(1 - i4)(1 +i4) = 1 +i4 -i4 - I^4 = 5

Answer 29

Pfft, what are you talking about Jhan, you're beyond smart

Answer 30

i^2 = -1

Answer 31

http://www.wolframalpha.com/input/?i=17+-+2+x+%2B+x%5E2%3D0

Answer 32

The solution is there, they put in parabola form for drawing. That's the problem when using the computer before you can do by hand. The computer only facilitates the work used be done by hand, slide rule, and calculators.

Answer 33

I'm old school, I'm 62 EE and gone through them all

Answer 34

@kenljw your solution is incorrect. you forgot to take a square root. don't believe me, try substituting your solution into the original equation.

Answer 35

specifically, this line is incorrect:

x = 1(1 +- sqrt(1 - 5(1^2)) = 1 +- i4

Answer 36

Your correct, that's why always show work so either you can check your arithmetic or someone else