Send i have no idea how to do this

1) x² - x - 12 = 0

Factor it

(x-4)(x+3) = 0

Study the factors

x-4 = 0 -> x =4
x+3 = 0 -> x = -3

2) t = x²

x⁴ = (x²)²

t² + 3t - 4 = 0

Use quadratic formula:

(-b±√b²-4ac)/2a

(-3±√9+16)/2

(-3±5)/2

t1 = 2/2 = 1
t2 = -8/2 = -4 (not possible, we choose t1)

So x² = 1

Result: x = ±1

3) Square both sides

2x-4 = (x-6)²

2x-4 = x²-12x+36

Take all terms to the left

2x - 4 - x² + 12x - 36 = 0

-x² + 14x - 40 = 0

x² - 14x + 40 = 0

Use quadratic formula:

(-b±√b²-4ac)/2a

(14±√196-160)/2

(14±6)/2

x1 = 10
x2 = 4

Let's check

√2(10) - 4 = 10 - 6
√20-4 = 4
√16 = 4
4 = 4 (OK)

√2(4) - 4 = 4 - 6

√8-4 = -2

√4 = -2

2 = -2 (NO)

So x = 10

4) Factor x²-9 = (x-3)(x+3)

So the common denominator would be (x-3)(x+3)



Cancel out the denominators...

x+3 ≠ 0 -> x ≠ -3
x-3 ≠ 0 -> x ≠ 3

Take all members to the left

x(x+3)+x(x-3)-2 = 0

x²+3x+x²-3x-2 = 0

2x² - 2 = 0

2x² = 2

x² = 1

x = ±1