x2-10 = 4x+11 x2-4x-10-11 = 0 x2-4x-21 = 0 (x+3)(x-7) = 0 x = -3 and x = 7
y = 4x-3
x3 + x2 + 4x + 4 = (x2 + 4)(x + 1)
X2+11x+11 = 7x+9 X2+11x-7x+11-9 = 0 x2+4x+2 = 0 Solve as a quadratic equation by using the quadratic equation formula or by completing the square: x = -2 + or - the square root of 2
y = x2 - 4x + 4 can be factored into y = (x-2)(x-2) The repeated factor is 2.
x2 + 4x + 3 = 0 (x + 1)(x + 3) = 0 x ∈ {-3, -1}
(2-r)e-rr
x2 + 4x = 1 x2 + 4x - 1 = 0 in the quadratic formula (b2+- Sqr(4ac))/2a, a = 1 b = 4 c = -1 so x = 8 +- i
x2-10 = 4x+11 x2-4x-10-11 = 0 x2-4x-21 = 0 (x+3)(x-7) = 0 x = -3 and x = 7
y = 4x-3
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
No.
x2 - 4x - 9 = 0 ∴ x2 - 4x = 9 ∴ x2 - 4x + 4 = 13 ∴ (x - 2)2 = 13 ∴ x - 2 = ±√13 ∴ x = 2 ± √13
Assuming that by writing that equation as a question, you mean to ask how to solve it for x, that can be done as follows: x2 + 4x - 6 = 0 x2 + 4x + 4 = 10 (x + 2)2 = 10 x + 2 = √20 x = -2 ± √20
x^2+4x+7
2x = x2 + 4x - 3 x2 + 2x - 3 = 0 (x - 1)(x - 2) = 0