Solve each equation by completing the square. a^2+16-66=-7
@wolf1728
ok,add 7 to both sides first :)
done
ok you have \(\sf \large a ^{2} + 16 - 66 + 7 = 0\)
can you say this? 16 - 66 + 7 = ?
yes I do.
nope...
16 - 66 + 7 = -43
so it's \( \sf \large a ^{2} - 43 = 0\)
you can solve it by using the formula,or you can solve it by using the laws of mathematics :)
the standard form of these equations are: \(\sf \Large Aa ^{2} + Ba + C = 0\)
and the formula is: \(\sf \LARGE a _{1,2} = \frac{ -b \ \pm \ \sqrt{b ^{2}-4ac}}{ 2a }\)
can you do it yourself?
so just use the quadratic formula to solve?
yes :)
okay thankyou. I think I can take it from there. :)
and do you want to know how to solve it by the laws of mathematics? besides that,do you want the proof of this formula?
what does that mean? sorry I'm not good at math...or english... to be honest.
oh no,no problem :) i'm not good at english too :) it mean: \(\ \sf \large Every \; equations \; can \; be \; solved \; by \; that \; formula \; or \; by \; the \; laws \; of \; mathematics \) what does "by the laws of mathematics" mean? i'll tell you :)
\( \sf a ^{2} - 43 = 0\) so can we say \( \sf a ^{2} = 43\) ?
yes we can.
@PFEH.1999
nice :) please solve this,then you can find you answer: \(\sf a ^{2} = 4\) solve for \( \sf a \)
a=2. right?
just 2?
\(\sf 2 ^{2} = 4\) it's right
but \(\sf(-2)^{2} = 4\) too :)
A quadratic function can be solved by • factoring • the quadratic formula OR • completing the square Knowing 1 of these methods will allow you to solve all quadratic equations but it seems algebra teachers want to show all 3 ways.
how can you solve\(\sf a^{2} = 4 \) by factoring ?:D
a² - 4 = 0
it's factoring ???
You would be factoring the difference of 2 squares
yes you can say this,but i have another meaning of "Factoring" ... and now @Mhm120 , got it? so \(\sf (-2)^{2} = 4\) too
Now that I think of it, trying to factor 3x² -17 +23 = 0 would be extremely difficult.
@wolf1728 , we're just confusing him . wait please.
Sure as a matter of fact, I think I'll just leave this topic.
no,no your quite right :) but he know knows the formula,and i wanted to tell him how to solve that by another way, and you're quite right too.just i said show your work after me :)
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