Example 1:
Find the roots of
, given that one root is the negative of the another.
Solution 1:
If the roots are a, b, and c, we have
, say. So, using Viete’s relations,


.
Hence,
and
. Hence, the roots are
.
Example 2:
Let
, with
such that a and
have the same sign. Show that the equation
cannot have both roots in the interval
.
Solution 2:
Let
be roots of the given quadratic equation. We have 
which in turn equals
.
If
both belong to
then each term of the sum will be negative, which is a contraction. Hence, the proof.
Example 3:
Consider all lines which meet the graph of
in four distinct points, say
. Then, show that
is independent of the line and find its value.
Solution 3:
Let
be any line which intersects the graph
at
, where
. Then,
are the roots of
.
(Note that
are distinct as
would imply
).
The above equation reduces to an equation of degree 4, namely,
. Hence, by Viete’s relations,
.
Example 4:
The product of two of the four roots of
is 24. Find k.
Solution 4:
Let the given equation be written as
, and let the roots of the equation be
with
. Now,
, so
. Also,

with
,
. Comparing coefficients of
and x we get
and
. This gives
. Comparing coefficients of
,
.
Example 5:
If
and
are roots of
, where p and q are integers with
, then show that
(i)
is an integer. (
)
(ii)
is an integer divisible by q. (
).
Solution 5:
Since
are the roots of
, we get
—– Equation A
—– Equation B
Note that
. For
, multiplying this equation by
, we get
. Similarly,
. Hence,
Eqn C
Also, for
,
Equation D
1.By equation A and D,
and
are both integers. Hence, by Eqn C, it follows by induction on n that
is an integer for
.
2. Since
, equation D shows that
. Further,
and so
. Hence, by C, it follows by induction on n that
for
.
Example 6:
Find all integers a such that the equation
has three integer roots.
Solution 6:
Let the integer roots of the given equation be
. Then,
,
. Let this be Equation I.
Hence,
. Let this be Equation II.
So,
and so the solution of Equation II are essentially the following:
—- Call this ***
—- Call this *****
Both these sets satisfy Equation I. Hence, the required values of a are
corresponding to the roots in (***) and (*****) respectively.
Example 7:
Find the remainder when
is divided by
.
Solution 7:
Dividing
by
, we get
, we get
.
Put
or
. Hence,

Using Binomial theorem, we get
, which in turn equals
Call this Equation @@@
Now, equating coefficients of
we get
,
,
.
Solving these equations, we get
,
, 
Hence, the remainder is

More algebraic stuff in the pipeline!!
Nalin Pithwa