To finish our problem,
we need to put
ax squared plus bx plus
c in the form a times x
minus h squared plus k.
Remember, we are doing
this so that x only
shows up in one place so
that we can solve for it.
To put ax squared plus bx
plus c in the form a times x
minus h squared plus k,
we need to figure out
the values of a, h, and
k so that ax squared
plus bx plus c equals a times
x minus h squared plus k,
just as we have been doing when
we knew the numbers a, b, and c
instead of leaving them generic.
So as with our examples, we
distribute the right hand side,
which gives us the
same thing as it always
has-- ax squared minus 2ahx
plus ah squared plus k.
And again like our
examples, we want
to find a, h, and k so our
coefficients are equal.
In particular, our
coefficient of x on the left
must be our coefficient
of x on the right.
So here we need
a equals a, which
shouldn't come as any surprise.
Although maybe we should have
made these different letters
to begin with.
And we probably
expect this anyways
from doing our examples.
We also need our coefficient of
x to be the same on both sides.
So b has to equal negative 2ah.
And finally, our constant has
to be the same on both sides.
Our constant c on the left has
to be our constant ah squared
plus k on the right.
Now, we want to solve for h.
We can do this by dividing by
negative 2a in this equation,
giving us b over
negative 2a equals
negative 2ah over negative
2a, which is just h.
And we can rewrite the
left as negative b over 2a.
So we have found that h
equals negative b over 2a.
Now for any polynomial ax
squared plus bx plus c,
we can find h as
negative b over 2a
without doing these steps
over and over again.
Now, plugging in what
we just found for m we
get that c is a times h squared,
which is negative b over 2a
squared plus k, which
is a times negative b
over 2a times negative
b over 2a plus k,
since that's what it
means to square something.
And we can simplify this,
because our a cancels
with our a.
And we can simplify this,
because the a in the top
cancels with one of
our a's in the bottom.
And then negative b times
negative b is b squared.
And 2 times 2a is 4a.
And we still have
plus k on the end.
But we want to solve this for k,
which we can do by subtracting
b squared over 4a.
This gives us c minus
b squared over 4a
equals b squared over 4a plus
k minus b squared over 4a.
And since b squared over
4a minus b squared over 4a
is just 0, we're just
left with k on the right.
So we found that k equals
c minus b squared over 4a.
We can, however, simplify
this a little bit.
Because we want to subtract
one number from the other,
but one's a fraction, we need
to get a common denominator.
To do this, we'll
multiply c by 4a
over 4a, Giving us 4ac over
4a minus b squared over 4a.
Now that we have a
common denominator,
we can combine
these to 4ac minus b
squared over that
common denominator, 4a.
So although they're very messy,
we have now found a, h, and k.
And we can conclude that we can
rewrite the polynomial we want
in the form ax squared plus
bx plus c equals a times
x minus negative
b over 2a, our h,
all squared plus 4ac minus
b squared over 4a, our k.
Let me point out that while
this has been very messy so far,
it will get better
and we'll have
a very beautiful
conclusion at the end.
