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anonymous
 one year ago
How would you use the quadratic formula to solve x^2+5x=2. Using compete sentences.
anonymous
 one year ago
How would you use the quadratic formula to solve x^2+5x=2. Using compete sentences.

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SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1maybe we can work the prob using math, and you do the sentences part alone?

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1when the equation is \(\LARGE\color{black}{ \color{blue}{a} x^2+ \color{magenta}{b}x+ \color{red}{c}=0 }\). you use the following formula: \(\normalsize\color{ slate }{\Huge{\bbox[5pt, lightcyan ,border:2px solid black ]{ \LARGE{x=~} \huge{ \frac{\color{magenta}{b} \pm\sqrt{ \color{magenta}{b} ^24 \color{blue}{a} \color{red}{c}}}{2 \color{blue}{a}} }~ }}}\) (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
 one year ago
Best ResponseYou've already chosen the best response.0So it will be x^2+5+2=0

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1what do you get after adding 2 to both sides?

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1\(\LARGE\color{black}{ \color{blue}{a} x^2+ \color{magenta}{b}x+ \color{red}{c}=0 }\) ` ... comparing ... ` \(\LARGE\color{black}{ \color{blue}{1} x^2+ \color{magenta}{5}x+ \color{red}{2}=0 }\)

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1Now, please plug in the a b c for me....

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0What do you mean by plug in?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I have no clue how to do this. I am just learning it

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.1she meant let a = 1, b = 5, and c = 2 we will have this equation \[x^2+5x+2=0\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Okay and then what do I do from there?

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.1Well, to solve this faster we need the discriminant \[b^24ac\] so let b = 5, a =1, and c =2

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.1umm I thought that was you?

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1well, 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
 one year ago
Best ResponseYou've already chosen the best response.1anyway plug in b =5, c =2, and a = 1 into the discriminant formula \[b^24ac\] @J_slate23

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.1as in replace the b with 5, replace the c with 2 and replace the a with 1

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.1yes but replace the = with 

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.1mhm so now what's 5^2 and what's 4(1)(2)

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.1\(\normalsize\color{ slate }{\Huge{\bbox[5pt, lightyellow ,border:2px solid black ]{ \LARGE{x=~} \huge{ \frac{\color{magenta}{5} \pm\sqrt{ \color{magenta}{5} ^24 \color{blue}{(1)} \color{red}{(2)}}}{2 \color{blue}{(1)}} }~ }}}\) and your equation is \(\LARGE\color{black}{ \color{blue}{1} x^2+ \color{magenta}{5}x+ \color{red}{2}=0 }\).

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.1yes so now what's 258

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.1yes! so for the bottom denominator what is 2(1) ?

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.1so 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
 one year ago
Best ResponseYou've already chosen the best response.1OH MAN I forgot the !

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.1\[\frac{5 +\sqrt{17}}{2}, \frac{5\sqrt{17}}{2}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Is that the answer? If so that was really easy

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.1yeah.. I just split the sign in the middle up. the latex for the entire quadratic equation with + and  together was a pain

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.1and 17 is not a perfect square... so we can't simplify .. not to mention 17 is a prime

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So the final answer is 5+sqrt17/2?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0That is easier then I expected

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Can you check my math for another one? @UskiDoll

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Solve x^2+4x12=0 by completing the square. X^2+4x12=0 Add 12 X^2+4x=12 (X+2)^24=12 (x+2)^2=16 (x+2)= 4 or 4 X= 6 or 6 is my answer

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Is that correct @UskiDoll

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.1\[x^2+4x12=0 \] \[x^2+4x=12 \] \[4 \times \frac{1}{2} = \frac{4}{2} = 2\] \[2^2 = 4\] \[x^2+4x+4=12+4 \] \[x^2+4x+4=16 \] \[(x+2)^2 = 16 \] square root on both sides \[\sqrt{(x+2)^2} = \sqrt{16}\] \[x+2=4, x+2=4\] \[x=2, x=6\]

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.1x+2 = 4 on your part is wrong. subtract 2 from both sides, but 6 is correct

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0plug it in b+_ b24ac \[\sqrt{b24ac}\] / 4a good luck

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.1The problem is already done ^_^
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