Question

X^2-6x-3=(x-a)^2-b, where a and b are constants

Answer 1

To find the values of "a" and "b", we can expand the right-hand side of the equation and equate it with the left-hand side:

x^2 - 6x - 3 = (x - a)^2 - b
x^2 - 6x - 3 = x^2 - 2ax + a^2 - b
0 = -2ax + a^2 - b + 6x + 3
0 = a^2 - 2ax + 6x + b + 3

Comparing the coefficients, we get:
a^2 - 2ax = 0
-2ax + 6x = 0

Solving for "a" and "b", we get:
a = 3
b = 9

So the equation becomes:
x^2 - 6x - 3 = (x - 3)^2 - 9

Answer 2

Below

Explanation:

This is an exercise in 'completing the square'

x^2 - 6x -3

take 1/2 of the x coefficient (6) square it and add it and subtract it

x^2 - 6x + 9 - 9 - 3 Now you can reduce the underlined portion

(x-3)^2 - 12 a = 3 b= -12 Done