Explanation:
1.1.1 x² - 6 x = 0
x × (x - 6) = 0
=> x = 0 , or, x - 6 = 0 , ie, x = 6.
1.1.2 x² - x - 5 = 0
x = [ 1 ± √(1+4×5)] /2 = [ 1 ± √21] /2
1.1.3. 3 √(x-2) = x - 6
9 (x - 2) = (x-6)² = x² -12 x + 36
=> x² -21 x + 54 = 0
=> x² - 18 x - 3 x + 54 = 0
=> x (x - 18) - 3 (x - 18) = 0
=> (x - 3) (x - 18) = 0
=> x = 3 or, 18.
1.2. y + x = 12
=> y = 12 - x.
x × y = 14 - 3 x
=> x (12 - x) = 14 - 3 x
=> 12 x - x² = 14 - 3 x
=>. x² - 15 x + 14 = 0
(x - 14)(x -1) = 0
=> x = 1 or 14.