Problem: Find the value of (a-b-c) if the zeroes of polynomial x^2-2x-3 are half of the zeroes of polynomial ax^2-bx-c.

Solution:

Step 1: Finding the zeroes of x^2-2x-3
We can find the zeroes of x^2-2x-3 by using the quadratic formula:

x = (-b ± √(b^2-4ac)) / 2a

Here, a=1, b=-2, c=-3. Substituting these values in the formula, we get:

x = (2 ± √(4+12)) / 2
x = (2 ± √16) / 2
x = 1 ± 2

So the zeroes of x^2-2x-3 are 3 and -1.

Step 2: Finding the zeroes of ax^2-bx-c
Let the zeroes of ax^2-bx-c be p and q. We are given that these zeroes are half of the zeroes of x^2-2x-3. So, we have:

p+q = (3+(-1))/2 = 1
pq = (3)(-1)/2 = -3/2

Step 3: Using Vieta's formulas
We can use Vieta's formulas to relate the coefficients of the polynomial ax^2-bx-c to its zeroes:

p+q = -b/a
pq = c/a

Substituting the values of p+q and pq from step 2, we get:

-b/a = 1
c/a = -3/2

Multiplying the first equation by a and substituting the second equation, we get:

-b = a
c = -3a/2

Step 4: Finding a-b-c
Substituting -b=a and c=-3a/2 in a-b-c, we get:

a-b-c = a-(-a)-(-3a/2) = 5a/2

So, the value of a-b-c is 5a/2.

Step 5: Finding the value of a
To find the value of a, we can use the fact that pq = -3/2. Substituting c=-3a/2 in this equation, we get:

-3a/2 * (1/a) = -3/2
-3/2 = -3/2

This equation is true for all values of a. So, there are infinitely many values of a that satisfy the given condition. However, we can still find a-b-c in terms of a:

a-b-c = 5a/2 = (5/2)(pq) = (5/2)(-3/2) = -15/4

Therefore, the value of a-b-c is -15/4.

From the given equation root are -1,3
but these are half of the zeroes of 2 and equation so zeroes of second equation are-2,6
a+b=4
ab=-12
×2-4×-12
a-b-c=1-4+12=9