x2+4x+4 = 25 x2+4x+4-25 = 0 x2+4x-21 = 0 (x+7)(x-3) = 0 x = -7 or x = 3
x2 + 4x = 0 ⇒ x(x + 4) = 0 ⇒ x = 0 or x = -4
x2 - 4x - 9 = 0 ∴ x2 - 4x = 9 ∴ x2 - 4x + 4 = 13 ∴ (x - 2)2 = 13 ∴ x - 2 = ±√13 ∴ x = 2 ± √13
x2 + 4x - 9 = 5x + 3 ∴ x2 - x - 12 = 0 ∴ (x + 3)(x - 4) = 0 ∴ x ∈ {-3, 4}
Rearrange: x2 - 4x + 4 = 0Factorise: (x - 2)(x - 2) = 0x = 2Check: 4 + 4 = 4 x 2? Sure does, job done.
x2+4x+4 = 25 x2+4x+4-25 = 0 x2+4x-21 = 0 (x+7)(x-3) = 0 x = -7 or x = 3
y = x2 - 4x + 4 = (x - 2)2 has one repeated root
x2 + 4x = 0 ⇒ x(x + 4) = 0 ⇒ x = 0 or x = -4
x2 - 4x - 9 = 0 ∴ x2 - 4x = 9 ∴ x2 - 4x + 4 = 13 ∴ (x - 2)2 = 13 ∴ x - 2 = ±√13 ∴ x = 2 ± √13
y = x2-4x+4 Since the highest degree term is 2, it must have 2 roots
x2 - 4x = 8 x2 - 4x + 4 = 8 + 4 (x - 2)2 = 12 x - 2 = +&- sq. root of 12 x = 2 +&- 2(sq. root of 3)
There are 2 roots to the equation x2-4x-32 equals 0; factored it is (x-8)(x+4); therefore the roots are 8 & -4.
x2 + 4x - 9 = 5x + 3 ∴ x2 - x - 12 = 0 ∴ (x + 3)(x - 4) = 0 ∴ x ∈ {-3, 4}
Rearrange: x2 - 4x + 4 = 0Factorise: (x - 2)(x - 2) = 0x = 2Check: 4 + 4 = 4 x 2? Sure does, job done.
Add 4 to both sides. To complete the square, take half the x coefficient into the bracket: (x + 4/2)2 = (x + 2)2 = x2 + 4x + 4 Comparing with the original, need to add 4 to both sides of x2 + 4x = 5: (x2 + 4x) + 4 = (5) + 4 ⇒ x2 + 4x + 4 = 9 ⇒ (x + 2)2 = 9
y = x2 - 4x + 4 can be factored into y = (x-2)(x-2) The repeated factor is 2.
x2+4x-9 = 5x+3 x2+4x-5x-9-3 = 0 x2-x-12 = 0 (x+3)(x-4) = 0 x = -3 or x = 4