**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