x2+7x+12
x2 + 7x + 9 = 3 ∴ x2 + 7x + 6 = 0 ∴ (x + 6)(x + 1) = 0 ∴ x ∈ {-6, -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
If: y = x2-7x+8 and y = -x2+9x-6 Then: x2-7x+8 = -x2+9x-6 So: 2x2-16x+14 = 0 => x2-8x+7 = 0 Therefore: x = 1 and x = 7 By substitution: x =1, y = 2 and x = 7, y = 8 Points of intersection: (1, 2) and (7, 8)
1
x2+7x+12
x2 + 7x = 2x2 + 6 x2 - 2x2 + 7x = 6 -x2 +7x - 6 = 0 or alternatively x2 - 7x + 6 = 0
(x + 3)(x + 4)
x2-7x+12 = 0 (x-3)(x-4) = 0x=3,4
=3x * * * * * The correct answer is: x2 + 7x + 12 = (x + 3)(x + 4).
x2-7x+12 (x-3)(x-4)
x2+7x+12=0usually this means to factor the equation.factored this equation would be...(x+3)(x+4)
The value of x2+7x+10 depends on the value of X. If x=1, then x2+7x+10 = 18 If x=2, then x2+7x+10 = 28 If x=3, then x2+7x+10 = 40 and so forth
x2 + 7x + 9 = 3 ∴ x2 + 7x + 6 = 0 ∴ (x + 6)(x + 1) = 0 ∴ x ∈ {-6, -1}
Do you mean x2+7x+12? If so then it is: (x+3)(x+4)
(x+3)(x+4)=x2+7x+12
x2 - 7x = 60 x2 - 7x - 60 = 0 (x - 12) (x + 5) = 0 x = 12 or x = -5