Question

How to solve this equation 441=40x+x^2 ????

Answer 1

\[441=40x+x ^{2}\]

Answer 2

start with
\[x^2+40x-441=0\] if you want to factor
otherwise it is probably easiest to complete the square
the problem was cooked up to factor because
\[x^2+40x-441=0\] factors as \[(x-9)(x+49)=0\] but it is not clear how you would know that

Answer 3

however, once you know it factors as
\[(x-9)(x+49)=0\] it is easy to solve
\[x-9=0\iff x=9\]\[x-49=x\iff x=-49\]

Answer 4

I don't understand, can you explain what your doing please?

Answer 5

if you did not see that it factors, you can always complete the square
starting with
\[x^2+40x=441\] you get
\[(x+20)^2=441+20^2=841\] and so
\[x+20=\pm\sqrt{841}=\pm29\] etc

Answer 6

ok lets go slow
first the factor method
you had \[441=40x+x ^{2}\]

Answer 7

subtracting \(441\) from both sides give
\[x^2+40x-441=0\] ok so far?

Answer 8

okay

Answer 9

then you need to factor this
it factors as \((x-9)(x+29)=0\)
this is because \(-9\times 49=-441\) and \(49-9=40\)

Answer 10

finding those factors is some magic, that is why i like completing the square more
but assuming we know that
\[x^2+40x-441=0\] is the same as
\[(x-9)(x+49)=0\] then we can solve easily

Answer 11

set each factor equal to zero and solve for \(x\) because if the product is zero, then one of the factors has to be zero
i wrote above \(x-9=0\iff x=9\) right?

Answer 12

yup!

Answer 13

Got it!

Answer 14

and also \(x+49=0\iff x=-49\)

Answer 15

Thanks you so much! Your awesome!!!!!

Answer 16

the hard part is factoring, i just said it, didn't show how to do it, but if it is clear, then so much the better
yw