Question

X^2-6x-3=(x-a)^2-b where a and b are constants.Find the values of a and b

Answer 1

a = 3, b = 12

Explanation:

x^2-6x-3=(x-a)^2-b

Lets expand the right side:

(x-a)^2-b = (x-a)(x-a)-b

(x-a)(x-a)-b = x^2 -2ax + a^2 - b

The left side is equal to the right, as per the original equation:

x^2-6x-3 = x^2 -2ax + (a^2 - b)

Compare each term:

1) x^2 = x^2 OK

2) -6x = -2ax; a = 3 so that -6x = -2(3)x

3) -3 = (a^2-b); a^2 = (b-3)

a =
`√(b-3)`

a is 3, from above

3 = 
`√(b-3)`

9 = b-3

b = 12