anonymous
  • anonymous
How would you use the quadratic formula to solve x^2+5x=-2. Using compete sentences.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
schrodinger
  • schrodinger
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anonymous
  • anonymous
@SolomonZelman
SolomonZelman
  • SolomonZelman
maybe we can work the prob using math, and you do the sentences part alone?
anonymous
  • anonymous
Okay

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SolomonZelman
  • SolomonZelman
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.)
anonymous
  • anonymous
So it will be x^2+5+2=0
SolomonZelman
  • SolomonZelman
what do you get after adding 2 to both sides?
SolomonZelman
  • SolomonZelman
yes
SolomonZelman
  • SolomonZelman
\(\LARGE\color{black}{ \color{blue}{a} x^2+ \color{magenta}{b}x+ \color{red}{c}=0 }\) ` ... comparing ... ` \(\LARGE\color{black}{ \color{blue}{1} x^2+ \color{magenta}{5}x+ \color{red}{2}=0 }\)
SolomonZelman
  • SolomonZelman
Now, please plug in the a b c for me....
anonymous
  • anonymous
What do you mean by plug in?
anonymous
  • anonymous
I have no clue how to do this. I am just learning it
UsukiDoll
  • UsukiDoll
she meant let a = 1, b = 5, and c = 2 we will have this equation \[x^2+5x+2=0\]
anonymous
  • anonymous
Okay and then what do I do from there?
UsukiDoll
  • UsukiDoll
Well, to solve this faster we need the discriminant \[b^2-4ac\] so let b = 5, a =1, and c =2
SolomonZelman
  • SolomonZelman
who is she?
UsukiDoll
  • UsukiDoll
umm I thought that was you?
SolomonZelman
  • SolomonZelman
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.
UsukiDoll
  • UsukiDoll
anyway plug in b =5, c =2, and a = 1 into the discriminant formula \[b^2-4ac\] @J_slate23
UsukiDoll
  • UsukiDoll
as in replace the b with 5, replace the c with 2 and replace the a with 1
anonymous
  • anonymous
So 5^2=4(1)(2)?
UsukiDoll
  • UsukiDoll
yes but replace the = with -
UsukiDoll
  • UsukiDoll
\[5^2-4(1)(2)\]
anonymous
  • anonymous
So 5^2-4(1)(2)?
UsukiDoll
  • UsukiDoll
mhm so now what's 5^2 and what's 4(1)(2)
SolomonZelman
  • SolomonZelman
\(\normalsize\color{ slate }{\Huge{\bbox[5pt, lightyellow ,border:2px solid black ]{ \LARGE{x=~} \huge{ \frac{-\color{magenta}{5} \pm\sqrt{ \color{magenta}{5} ^2-4 \color{blue}{(1)} \color{red}{(2)}}}{2 \color{blue}{(1)}} }~ }}}\) and your equation is \(\LARGE\color{black}{ \color{blue}{1} x^2+ \color{magenta}{5}x+ \color{red}{2}=0 }\).
anonymous
  • anonymous
25 and 8
UsukiDoll
  • UsukiDoll
yes so now what's 25-8
anonymous
  • anonymous
17
UsukiDoll
  • UsukiDoll
yes! so for the bottom denominator what is 2(1) ?
anonymous
  • anonymous
2
UsukiDoll
  • UsukiDoll
yes
UsukiDoll
  • UsukiDoll
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
UsukiDoll
  • UsukiDoll
OH MAN I forgot the -!
UsukiDoll
  • UsukiDoll
\[\frac{-5 +\sqrt{17}}{2}, \frac{-5-\sqrt{17}}{2}\]
anonymous
  • anonymous
Is that the answer? If so that was really easy
UsukiDoll
  • UsukiDoll
yeah.. I just split the sign in the middle up. the latex for the entire quadratic equation with + and - together was a pain
UsukiDoll
  • UsukiDoll
and 17 is not a perfect square... so we can't simplify .. not to mention 17 is a prime
anonymous
  • anonymous
So the final answer is -5+-sqrt17/2?
UsukiDoll
  • UsukiDoll
yeah
anonymous
  • anonymous
Thank you!
anonymous
  • anonymous
That is easier then I expected
anonymous
  • anonymous
Can you check my math for another one? @UskiDoll
UsukiDoll
  • UsukiDoll
sure
anonymous
  • anonymous
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
anonymous
  • anonymous
Is that correct @UskiDoll
UsukiDoll
  • UsukiDoll
\[x^2+4x-12=0 \] \[x^2+4x=12 \] \[4 \times \frac{1}{2} = \frac{4}{2} = 2\] \[2^2 = 4\] \[x^2+4x+4=12+4 \] \[x^2+4x+4=16 \] \[(x+2)^2 = 16 \] square root on both sides \[\sqrt{(x+2)^2} = \sqrt{16}\] \[x+2=4, x+2=-4\] \[x=2, x=-6\]
UsukiDoll
  • UsukiDoll
x+2 = 4 on your part is wrong. subtract 2 from both sides, but -6 is correct
anonymous
  • anonymous
Okay. Thank you
anonymous
  • anonymous
plug it in -b+_ b2-4ac \[\sqrt{b2-4ac}\] / 4a good luck
UsukiDoll
  • UsukiDoll
The problem is already done ^_^

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