EXPLANATION

EXPLANATION

You have already seen that zero of a linear polynomial ax + b is . We will now try to answer the question raised in Section 2.1 regarding the relationship between zeroes and coefficients of a quadratic polynomial. For this, let us take a quadratic polynomial, say p(x) = 2x² – 8x + 6. In Class IX, you have learnt how to factorise quadratic polynomials by splitting the middle term. So, here we need to split the middle term ‘– 8x’ as a sum of two terms, whose product is 6 × 2x² = 12x². So, we write

2x² – 8x + 6 = 2x² – 6x – 2x + 6 = 2x(x – 3) – 2(x – 3)

= (2x – 2)(x – 3) = 2(x – 1)(x – 3)

So, the value of p(x) = 2x² – 8x + 6 is zero when x – 1 = 0 or x – 3 = 0, i.e., when x = 1 or x = 3. So, the zeroes of 2x² – 8x + 6 are 1 and 3. Observe that :

Sum of its zeroes = 

Product of its zeroes = 

In general, if * and * are the zeroes of the quadratic polynomial p(x) = ax² + bx + c, a  0, then you know that x –  and x –  are the factors of p(x). Therefore,

ax² + bx + c = k(x – ) (x – ), where k is a constant

= k[x² – ( +)x +]

= kx² – k(+ )x + k

Comparing the coefficients of x², x and constant terms on both the sides, we get

a = k, b = – k( and c=k
This gives 

i.e., sum of zeroes = 

product of zeroes = 

Let us consider some examples.

Example 2 and 3

Example 2 and 3

EXAMPLE 2

Find the zeroes of the quadratic polynomial x² + 7x + 10, and verify the relationship between the zeroes and the coefficients.

SOLUTION:

We have x² + 7x + 10 = (x + 2)(x + 5)

So, the value of x² + 7x + 10 is zero when x + 2 = 0 or x + 5 = 0, i.e., when x = – 2 or x = –5. Therefore, the zeroes of x² + 7x + 10 are – 2 and – 5. Now,

Sum of zeroes = 

Product of zeroes = 

EXAMPLE 3

Find the zeroes of the polynomial x² – 3 and verify the relationship between the zeroes and the coefficients.

SOLUTION

Recall the identity a² – b² = (a – b)(a + b). Using it, we can write:

So, the value of x² – 3 is zero when x = or x =.

Therefore, the zeroes of x² – 3 are  and .

Now

Sum of zeroes = 

Product of zeroes = 

EXAMPLE 4 and 5

EXAMPLE 4 and 5

Find a quadratic polynomial, the sum and product of whose zeroes are – 3 and 2, respectively.

Solution:

Let the quadratic polynomial be ax² + bx + c, and its zeroes be  and .

We have 

and 

If a = 1, then b = 3 and c = 2.

So, one quadratic polynomial which fits the given conditions is x² + 3x + 2.

You can check that any other quadratic polynomial that fits these conditions will be of the form k(x² + 3x + 2), where k is real.

Let us now look at cubic polynomials. Do you think a similar relation holds between the zeroes of a cubic polynomial and its coefficients?

Let us consider p(x) = 2x³ – 5x² – 14x + 8.

You can check that p(x) = 0 for x = 4, – 2, Since p(x) can have atmost three zeroes, these are the zeores of 2x³ – 5x² – 14x + 8. Now,

sum of the zeroes = 

product of the zeroes = 

However, there is one more relationship here. Consider the sum of the products of the zeroes taken two at a time. We have

.

In general, it can be proved that if  are the zeroes of the cubic polynomial ax³ + bx² + cx + d, then

Let us consider an example.

EXAMPLE 5

Verify that 3, –1, are the zeroes of the cubic polynomial p(x) = 3x³ – 5x² – 11x – 3, and then verify the relationship between the zeroes and the coefficients.

SOLUTION:

Comparing the given polynomial with ax³ + bx² + cx + d, we get

a = 3, b = – 5, c = –11, d = – 3. Further

p(3) = 3 × 3³ – (5 × 3²) – (11 × 3) – 3 = 81 – 45 – 33 – 3 = 0,

p(–1) = 3 × (–1)³ – 5 × (–1)² – 11 × (–1) – 3 = –3 – 5 + 11 – 3 = 0,

Therefore, 3, –1 and  are the zeroes of 3x³ – 5x² – 11x – 3.

So, we take = 3,  = –1 and .

Now