Question

If the zeroes of polynomial:1x-2x-3 are half of the zeroes of polynomial ax+BC+c. Find(a-b-c)?

Answer 1

Given ,
The roots or zeroes of x² -2x - 3 are half the zeroes of the polynomial ax²+bx+c .

Let us first find the zeroes, of x² - 2x - 3
⇒ x² - 3x + x - 3
⇒ x ( x - 3 ) + 1 ( x - 3 )
⇒ ( x - 3 ) ( x + 1 )

Zeroes of the polynomial :
( x - 3 ) ( x + 1 ) = 0
x = 3 , x = -1

Let the zeroes of the polynomial ax² + bx + c are p, q

According to the question,
p/2 = 3 ⇒ p = 3×2 = 6
q/2 = -1 ⇒ q = 2 ( -1 ) = -2 .

Now, Sum of roots = 6 + ( -2 ) = 4
Product of roots = 6 ( -2) = -12

The required quadratic polynomial = x² - 4x - 12 .

( x - 6 ) ( x + 2 )
= x ( x + 2 ) -6 ( x + 2 )
= x² + 2x -6x -12
= x² - 4x -12

Comparing this with ax² + bx + c We get,
a = 1 , b = -4 , c = -12

Now,
( a - b - c )
= 1 - ( -4 ) - ( -12 )
= 1 + 4 + 12
= 17

Therefore, The value of a - b - c is 17

Answer 2

Hello Dear.

There is the little mistakes in the Question. The Polynomial will be x² - 2x - 3 and the another polynomial will be ax² + bx + c.

Solutions ⇒ 

Let the Zeroes of the Polynomial ax² + bx + c be m and n.

For the Polynomial x² - 2x - 3

Splitting the Middle term,

x² - 3x + x - 3
⇒ x(x - 3) + 1(x - 3)
⇒ (x - 3)(x + 1)

∴ Zeroes of the Polynomial ⇒
(x - 3)(x + 1) = 0
⇒ x = 3   and   x = -1


According to the Question,

 Zeroes of the x² - 2x -3 are equals to the half of the zeroes of the ax² + bx + c

∴ m/2 = 3
   m = 6

∴ n/2 = -1
  n = -2

Thus, the Zeroes of the Polynomial ax² + bx + c  are 6 and -2.

Now, Let us Find the Quadratic Equation of these Zeroes.

∴ x = 6 and x = -2
⇒ (x - 6) = 0 and (x + 2) = 0
⇒ (x - 6)(x + 2) = 0
⇒ x² + 2x - 6x - 12 = 0
⇒ x² - 4x - 12 = 0

Now, On comparing this proved quadratic equation with the ax² + bx + c = 0,
We get,

a = 1
b = -4
c = -12

∴ a - b - c = 1 - (-4) - (-12)
                = 1 + 4 + 12
                = 5 + 12
                = 17



∴ The Final Value of (a - b - c) is 17.


Hope it helps.