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what value do you think we should add?
I am not sure 9
x^2+18x = x^2 + 2*9*x +0 here's the formula: a^2 + 2ab + b^2 = (a+b)^2 so, you can see a=x, b=9. you are missing b^2, to complete square you'll need b^2, which is 9^2=81 so add 81 to each side.
Is That it
try it, then you tell me the result.
9 is a good number to use, but it is naked the way it is; square it and were good :)
I am really confused should I square the complete equation?
you can't square the whole equation. you should end up with something like this (x+a)^2+b=0 a and be can be positive or negative.
except for some cases where you can just put them together nicely, of course.
i had a similar problem and don't understand how to do it and how they came up with the answer it was x^2+16x=-7 the answer was 64
try to put the left side into this form a^2 + 2ab + b^2. add something to make b^2, remember, you add something, you have to subtract it
a=1 b=18 c=-8
I know that I am having problems solving this because I can't find an example to use
b^2 = (2ab/2a) *b
ok, I'll do you this one example. x^2+16x=-7 x^2+16x+7=0 x^2+2*x*8+7=0 compare this to a^2 + 2*a*b + b^2=(a+b)^2 you can see clearly that a=x b=8 so if b=8, b^2=64, you need to add 64, but you add 64, you need to subtract 64 then equation becomes x^2 + 2*x*8 + 64 - 64 + 7=0 (a+8)^2 + (-64+7) = 0 (a+8)^2 - 57 = 0 there's the answer
18/2(1) = 9*9 = 81
(a+9)^2-73=0 So I would add 81 to both sides
??? I dont get it.
x^2+18x=-8 x^2+2*x*9+81-81+8=0 (a+9)^2+(-81+8)=0 (a+9)^2-73=0 I would add 81 to both sides
exactly. that's right
Thank you now I have an example to go by
5x^2-3=0 no solution is that correct?
thank you got to go let me know if that is right
you mean to complete square, no, but there are solutions, add 3 to both sides, then divide both sides by 5, you'll get x^2=3/5=(sqrt(3/5))^2=(-sqrt(3/5))^2 so x=sqrt(3/5) or x=-sqrt(3/5)
thanks i figured it out square root property
ah that's good, good luck with ur maths