So i think you have to solve the equation by completing the square..please i really need help..:/ a^2-10a-39=0
do you know what completing the square is?
Yes i do, i mean we just started this but im still a little hung up on the process
ok, so first, add 39 to both sides, tell me what you get
Its 39?
no, i mean do this: a^2-10a-39+39=39. Simplify that for me
a^2-10a=39
@HaileyD here is a suggestion \[a ^{2} +13a-3a-39\] = 0 try next ...... :)
right, so now the goal is to get the left side to become a PERFECT SQUARE TRINOMIAL
do you know how to make a^2-10a into a perfect square trinomial?
(b/2)^2? which was (10/2)^2 and i got 25 then i added 25 to both sides and thats where i got lost
GREAT! so add that to both sides, simplify and write what you got
Is it (a-8)^2=64?
no, i want you write the non-factored version
ok, i'll write it: a^2-10a+25=39+25. Simplify that if you can. Tell me if you need help
\[(a ^{2}-13a) + (3a-39)=0\] \[a(a-13) + 3(a-13) =0\] \[so , (a+3) (a-13) =o \] \[\in fact (a+3) =0 Or (a-13) = 0 is must\] so , answers are ... a= (-3) Or a= 13. :)
hey hitswild, dont give direct answers
and your method is incorrect
a^2-10a+25=64?
correct, now you need to factor the left side into a perfect square
\(\large\color{black}{ \displaystyle a^2-10a-39=0 }\) \(\large\color{black}{ \displaystyle a^2-10a=39 }\) \(\large\color{black}{ \displaystyle a^2-10a+25=39+25 }\) \(\large\color{black}{ \displaystyle(a-5)^2=64 }\) and on...
@Rizags O'oo is that so? I'm sorry
@SolomonZelman has it factored correctly. Can you take it from his last step?
Okay thanks ill see what i get...
I got it a=13,-3
@HaileyD - I'm not familiar with English so I don't have a proper Idea about the meaning of "the equation by completing square" but I'm pretty sure the way I got the answers.. pls check the correctness with your fellows.. And the laast step giver by @SolomonZelman is wrong it must be as "\[(a ^{2}-5^{2}) = \sqrt{64} = 8\]" \ ( And keep in mind that [(a ^{2}-5^{2}) = (a-5) (a+5) \] ) :) good luck try the best and I feel like I 've given all the steps and answers directly and It mustn't. I had to let you think and try it yourself. :(
I dont know if this means or changes anything but im in common core math and it sucks, but the way @Rizags did it looked more familiar to me
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