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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
 one year ago
 one year 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
 one year ago
 one year ago

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ChristianGeekBest ResponseYou've already chosen the best response.0
You need to change the signs...(x + 2)(x 12) = x^2  10x 24!
 one year ago

kirk.freedmanBest ResponseYou've already chosen the best response.1
OHH 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.
 one year ago

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

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

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

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

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

HeroBest ResponseYou've already chosen the best response.1
The 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.
 one year ago

kirk.freedmanBest ResponseYou've already chosen the best response.1
Thanks so much, I get it:)
 one year ago
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