anonymous
  • anonymous
find the solution to the quadratic equation x^2 + x = 3x - 5
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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whpalmer4
  • whpalmer4
it's right there in front of you! did you mean find the solutions to the quadratic equation?
anonymous
  • anonymous
oops , yes the solution! find the solution!
anonymous
  • anonymous
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

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More answers

whpalmer4
  • whpalmer4
This is also pretty easy to solve by completing the square, if you prefer.
anonymous
  • anonymous
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
anonymous
  • anonymous
someone clarify pleeeeasee
Luigi0210
  • Luigi0210
move everything to one side and then plug in values to the quadratic formula
anonymous
  • anonymous
my answer is D
whpalmer4
  • whpalmer4
D is correct
dan815
  • dan815
luigi!
dan815
  • dan815
nice score! :)
Jhannybean
  • Jhannybean
Using @whpalmer4's way, let's try completing the square. \[\large x^2 +x =3x -5\]\[\large x^2 + x - 3x = -5\]\[\large x^2 -2x = -5\]\[ \quad \large c=(\frac{-2}{2})^2=1\]\[\large x^2 - 2x +1 = -5+1\]\[\large (x-1)^2 = -4\]\[\large \sqrt{(x-1)^2}=\sqrt{-4}\]\[\large x-1 = \pm 2i\]\[\large x= 1 \pm2i \implies x- 2i, x+2i\]
whpalmer4
  • whpalmer4
I would have stopped at \(x=1\pm2i\), not sure what the rightarrow is supposed to indicate...
Luigi0210
  • Luigi0210
you will all surpass me eventually, dan :P
Jhannybean
  • Jhannybean
"implies"
dan815
  • dan815
no i wont i will wait for you level then i will lvl
whpalmer4
  • whpalmer4
okay, but I don't see how it implies what you have to the right of the arrow :-)
Jhannybean
  • Jhannybean
@dan815 mine is better. ;) lolol.
dan815
  • dan815
69 is that akward phase no one wants to be in
dan815
  • dan815
its like being 18 before u can drink
Luigi0210
  • Luigi0210
well you're gonna have to wait awhile then :P
Jhannybean
  • Jhannybean
what i mean is if you break it down "x pm 2i" means x=1+ 2i and x= 1-2i lol.
KenLJW
  • KenLJW
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
Luigi0210
  • Luigi0210
@dan815 watch, Jhanny will even pass me.. then she'll be your competition :P
Jhannybean
  • Jhannybean
I don't feel smart enough!
whpalmer4
  • whpalmer4
@kenljw but what about that pesky detail that \[(x-1+4i)(x-1-4i) = x^2-2x+17\]
whpalmer4
  • whpalmer4
sqrt(1-5(1^2) = 2i, not 4i
KenLJW
  • KenLJW
(1 - i4)(1 +i4) = 1 +i4 -i4 - I^4 = 5
Luigi0210
  • Luigi0210
Pfft, what are you talking about Jhan, you're beyond smart
KenLJW
  • KenLJW
i^2 = -1
whpalmer4
  • whpalmer4
http://www.wolframalpha.com/input/?i=17+-+2+x+%2B+x%5E2%3D0
KenLJW
  • KenLJW
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.
KenLJW
  • KenLJW
I'm old school, I'm 62 EE and gone through them all
whpalmer4
  • whpalmer4
@kenljw your solution is incorrect. you forgot to take a square root. don't believe me, try substituting your solution into the original equation.
whpalmer4
  • whpalmer4
specifically, this line is incorrect: x = 1(1 +- sqrt(1 - 5(1^2)) = 1 +- i4
KenLJW
  • KenLJW
Your correct, that's why always show work so either you can check your arithmetic or someone else

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