-x2 + x + 42 = -[x2 - x - 42] = -[x2 - 7x + 6x - 42] = -[x*(x - 7) + 6*(x - 7)] = -(x - 7)*(x + 6)
-x2 + x + 42= -x2 + 7x - 6x + 42= 7x - x2 + 42 - 6x= x(7 - x) + 6(7 - x)= (7 - x)(x + 6)
42 + x2 = 13x ∴ x2 - 13x + 42 = 0 ∴ (x - 6)(x - 7) = 0 ∴ x ∈ {6, 7}
This is a quadratic equation question which will have two answers: 2x2+42 = x2+13x 2x2-x2-13x+42 = 0 x2-13x+42 = 0 Factorising the equation gives you: (x-6)(x-7) = 0 Therefore: x = 6 or x = 7
If you're looking to solve for x, you can do so as follows: x2 + 12x - 6 = 0 x2 + 12x + 36 = 42 (x + 6)2 = 42 x + 6 = ± √42 x = -6 ± √42
-x2 + x + 42 = -[x2 - x - 42] = -[x2 - 7x + 6x - 42] = -[x*(x - 7) + 6*(x - 7)] = -(x - 7)*(x + 6)
-x2 + x + 42= -x2 + 7x - 6x + 42= 7x - x2 + 42 - 6x= x(7 - x) + 6(7 - x)= (7 - x)(x + 6)
(-x-6)(x-7)
42 + x2 = 13x ∴ x2 - 13x + 42 = 0 ∴ (x - 6)(x - 7) = 0 ∴ x ∈ {6, 7}
This is a quadratic equation question which will have two answers: 2x2+42 = x2+13x 2x2-x2-13x+42 = 0 x2-13x+42 = 0 Factorising the equation gives you: (x-6)(x-7) = 0 Therefore: x = 6 or x = 7
If you're looking to solve for x, you can do so as follows: x2 + 12x - 6 = 0 x2 + 12x + 36 = 42 (x + 6)2 = 42 x + 6 = ± √42 x = -6 ± √42
It is: (x+7)(x-6) when factored
(x + 7)(x - 6)
44
No, as it has two factors, (x - 6)(x + 7)
x3 + 13x2 + 42x = x(x2 + 13x + 42) = x(x2 + 6x + 7x + 42) = x[x(x + 6) + 7(x + 6)] = x(x + 7)(x + 6)
Given that 52 - x2 + x = 10, then -52 + x2 - x + 10 = x2 - x - 42 = (x - 7)(x + 6) = 0. Therefore, x = 7 or -6.