Given that $x^2 - 6x + 1 = (x - a)^2 - b$ for all values of $x$, (i) find the value of $a$ and the value of $b$ - Edexcel - GCSE Maths - Question 20 - 2019 - Paper 1

To solve the equation

$x^2 - 6x + 1 = (x - a)^2 - b$

we first expand the right-hand side.

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

Setting the two sides equal gives us:

$x^2 - 6x + 1 = x^2 - 2ax + (a^2 - b)$

Comparing coefficients for $x$ terms:

$-6 = -2a$

Solving for $a$:

$a = 3$

Next, comparing the constant terms:

$1 = a^2 - b$

Substituting $a = 3$ into the equation yields:

$1 = 3^2 - b$

which simplifies to:

$1 = 9 - b$

Thus,

$b = 8$

Therefore, the values are

$a = 3$ and $b = 8$.