4x2 - 2x = 0
(2x + 3)(2x + 3)
(2x - 3)*(4x2 + 6x + 9) = 0
-4x2 + 6x + 40 = 0 ∴ -2x2 + 3x + 20 = 0 ∴ -2x2 - 8x + 5x + 20 = 0 ∴ -2x(x + 4) + 5(x + 4) = 0 ∴ (5 - 2x)(x + 4) = 0 ∴ x ∈ {5/2, -4}
5x2-2x+16=4x2+6x ,or x2-8x+16=0, or x2-2(x)(4)+(4)2=0, or (x-4)2=0, or (x-4)(x-4)=0, or x=4
4x2 - 2x = 0
Assuming that 4x2+2x = 0: 2x (2x+1) = 0 2x = 0 or 2x+1 = 0 x = 0 or -1/2
4x2 - 2x - 3 = 0 (2x + 1)(2x - 3) = 0 x = -(1/2) or (3/2)
4x2-4x-3 = 0 (2x-3)(2x+1) = 0 x = 3/2 or x = -1/2
(2x + 3)(2x + 3)
(2x - 3)*(4x2 + 6x + 9) = 0
4x2 - 4x + 1 = 0 => (2x - 1)(2x - 1) = 0 => (2x - 1)2 = 0 There is one solution: x = 1/2. It is a repeated root of the equation.
4x2+2x+1
Given, 4x2 - 4x + 3 = 0; then, 4x2 - 4x + 1 = (2x - 1)2 = -2, 2x - 1 = √-2 = ±i√2, and 2x = 1 ± i√2; therefore, x = ½(1 ± i√2).
-4x2 + 6x + 40 = 0 ∴ -2x2 + 3x + 20 = 0 ∴ -2x2 - 8x + 5x + 20 = 0 ∴ -2x(x + 4) + 5(x + 4) = 0 ∴ (5 - 2x)(x + 4) = 0 ∴ x ∈ {5/2, -4}
5x2-2x+16=4x2+6x ,or x2-8x+16=0, or x2-2(x)(4)+(4)2=0, or (x-4)2=0, or (x-4)(x-4)=0, or x=4
Discriminant = b2-4ac = (-2)2-(4*4*4) = -60