If the zeroes of the polynomial $$x^2 - ax + b$$ are 3 and 4, then ‘a’ and ‘b’ are respectively equal to

Given: The zeroes of the polynomial $$x^2 - ax + b$$ are 3 and 4.

To find: The values of a and b.

Answer:

Let p(x) be $$x^2 - ax + b.$$

Let's find the value of the equation by substituting x with 3.

p(3) = $$ (3^2 - 3a + b)$$

p(3) = 9 - 3a + b

9 - 3a + b = 0

-3a + b = -9

b = -9 + 3a → (Equation 1)

Now, let us find the value when x = 4.

p(4) = $$ (4^2 - 4a + b)$$

-4a + b = -16

b = -16 + 4a

Substituting (Equation 1) in place of b,

-9 + 3a = -16 + 4a

-9 + 16 = 4a - 3a

7 = a

a = 7

Now, substituting the value of a in b = -9 + 3a.

b = -9 + 3×7

b = -9 + 21

b = 12

Therefore, a = 7 and b = 12.