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kirk.freedman
 3 years ago
Solve the quadratic expression by completing the square.
x2 + 10x – 17 = 7
I did the problem but when I plug my answers into the original equation it doesn't work. this is what i got.
x^2+10x24=0
(x+2)(x12)=0
x=2,12
kirk.freedman
 3 years ago
Solve the quadratic expression by completing the square. x2 + 10x – 17 = 7 I did the problem but when I plug my answers into the original equation it doesn't work. this is what i got. x^2+10x24=0 (x+2)(x12)=0 x=2,12

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ChristianGeek
 3 years ago
Best ResponseYou've already chosen the best response.0You need to change the signs...(x + 2)(x 12) = x^2  10x 24!

kirk.freedman
 3 years ago
Best ResponseYou've already chosen the best response.1OHH I see, and wait do I have to use some other method to solve this, since it says to solve by completing the square. I have to show my work so I have to make sure I do the problem the way it says or I will get it wrong.

ChristianGeek
 3 years ago
Best ResponseYou've already chosen the best response.0Good point...you do need to use a different method. See this link: http://www.purplemath.com/modules/sqrquad.htm

Hero
 3 years ago
Best ResponseYou've already chosen the best response.1Yes, he should learn how to complete the square because x^2  10x  24 is not even the right polynomial to use.

kirk.freedman
 3 years ago
Best ResponseYou've already chosen the best response.1Can I please get some help on how to do this the right way? I'm confused

kirk.freedman
 3 years ago
Best ResponseYou've already chosen the best response.1I think I probably know how to complete the square, I just need to rejog my memory a bit

Hero
 3 years ago
Best ResponseYou've already chosen the best response.1\[x^2+10x24=0 \\x^2 + 10x = 24 \\x^2 + 10x + 25 = 24 + 25 \\(x + 5)^2 = 49 \\x + 5 = \pm 7 \\x = \pm7  5 \\x = 2 \\x = 12\]

Hero
 3 years ago
Best ResponseYou've already chosen the best response.1The trick is to remember to get whatever you're working with to the form \(ax^2 + bx = c\) Then afterwards, add \(\left(\frac{b}{2}\right)^2\) to both sides.

kirk.freedman
 3 years ago
Best ResponseYou've already chosen the best response.1Thanks so much, I get it:)
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