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Someone please be awesome and help me please :(

Someone please be awesome and help me please :(-example-1
User Ulphat
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Answer:


(x+(b)/(2a))^2+((4ac)/(4a^2)-(b^2)/(4a^2))=0


(x+(b)/(2a))^2=(b^2-4ac)/(4a^2)


x=(-b)/(2a) \pm (√(b^2-4ac))/(2a)

Explanation:


x^2+(b)/(a)x+(c)/(a)=0

They wanted to complete the square so they took the thing in front of x and divided by 2 then squared. Whatever you add in, you must take out.


x^2+(b)/(a)x+((b)/(2a))^2+(c)/(a)-((b)/(2a))^2=0

Now we are read to write that one part (the first three terms together) as a square:


(x+(b)/(2a))^2+(c)/(a)-((b)/(2a))^2=0

I don't see this but what happens if we find a common denominator for those 2 terms after the square. (b/2a)^2=b^2/4a^2 so we need to multiply that one fraction by 4a/4a.


(x+(b)/(2a))^2+(4ac)/(4a^2)-(b^2)/(4a^2)=0

They put it in ( )


(x+(b)/(2a))^2+((4ac)/(4a^2)-(b^2)/(4a^2))=0

I'm going to go ahead and combine those fractions now:


(x+(b)/(2a))^2+((-b^2+4ac)/(4a^2))=0

I'm going to factor out a -1 in the second term ( the one in the second ( ) ):


(x+(b)/(2a))^2-((b^2-4ac)/(4a^2))=0

Now I'm going to add (b^2-4ac)/(4a^2) on both sides:


(x+(b)/(2a))^2=(b^2-4ac)/(4a^2)

I'm going to square root both sides to rid of the square on the x+b/(2a) part:


x+(b)/(2a)=\pm \sqrt{(b^2-4ac)/(4a^2)}


x+(b)/(2a)=\pm (√(b^2-4ac))/(2a)

Now subtract b/(2a) on both sides:


x=(-b)/(2a) \pm (√(b^2-4ac))/(2a)

Combine the fractions (they have the same denominator):


x=(-b \pm √(b^2-4ac))/(2a)

User Hugo Maxwell
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