(i) Given x2 - 3x - 10 = 0
x2-x + 2x - 10 = 0
x(x - ) + 2(x - ) = 0
(x - )(x + 2) = 0
x - = 0 or x + 2 = 0
x = or x = -2
x = or -2 are the roots of the given quadratic equation.
2x 2 + 4x - x - 6 = 0
2x(x + 2) - (x + 2) = 0
(x + 2)(2x - ) = 0
(x + 2) or 2x - = 0
x = -2 or 2x =
x = -2 or√2x2+x + 2x + 5√2 = 0
x(√2x + ) + √(√2x + ) = 0
(√2x + )(x + √)= 0
√2x + = 0 or x + √= 0
√2x = - or x = -√
x =16x2-x - 4x + 1 = 0
x(4x - 1)-(4x - 1) = 0
(x - )(4x - 1) - 0
x - = 0
x =
x =100x2 - 10x - 10x + 1 = 0
10x(10x - 1) - 1(10x - 1) = 0
(10x - 1)(10x - 1) = 0
10x - 1 = 0
10x = 1
x =(vi)Given x(x + 4) = 12
x2 + 4x = 12
x2 + 4x - 12 = 0
x2+ x - 2x - 12 = 0
x(x + ) - 2(x+) = 0
(x + )(x - 2) = 0
x + = 0 or x - 2 = 0
x = - or 2 are the roots of the given quadratic equation.
3x2-x - 2x + 2 = 0
x(x - 1)-(x - 1) = 0
(x - 1)(x - ) = 0
x - 1 = 0 or x- = 0
x = 1 or are the roots of the quadratic equation.x2 - 2x - 3 = 0
x2-x + x - 3 = 0
x(x - )+ 1(x -) = 0
(x - )(x + 1) = 0
x - = 0 or x + 1 = 0
x = or -1 are the roots of the given equation.
Take (x - 4) = a, then the given quadratic equation reduces to 3a2- 5a = 12
3a2- 5a - 12 = 0
3a2- 9a + a - 12 = 0
3a(a - 3) + (a - 3) = 0
a - 3 = 0 or 3a + = 0
a = 3 or a =