The logic of these factored quadratic equations is:-
If A x B = 0 then
either A = 0 or B = 0, (or both).
a) (x+5)(x-2) = 0
Either (x+5)= 0 in which case x = -5
or (x-2) = 0, in which case x = +2
These are the two possible answers or solutions
b) x^2 - 36 = 0
One of the first things you learn in algebra is
the difference of two squares
a^2 - b^2 = (a + b)(a - b)
This helps to factorise the example given as
x^2 - 36 = (x + 6)(x - 6) = 0
Using the same logic as before,
either x = -6 or x = +6
Your teacher is trying to emphasise that equations
with x^2 as the high power, (called quadratics),
have two solutions.
It is easy to miss one. If I ask what is x, if x squared is 4,
the answer usually given is 2
and the other answer, -2 can get forgotten.
Factorising helps to get both solutions.
c) 3x^2 = 54
x^2 = 18 = 3 * 3 * 2 = [3* SQRT of(2)]^2
Rearranging, this is another difference of two squares
x^2 - [3* SQRT of(2)]^2 = 0
[x + 3SQRT(2)] [x - 3SQRT(2)] = 0
x = -3SQRT(2) or x = +3SQRT(2)
since SQRT(2) = 1.414....
x = -4.24 or x = +4.24
(rounded to the nearest hundredth)
Regards - Ian
