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 + 6 = (x + 6) (x + 1)
x2 + 7x - 8 = (x + 8)(x - 1)
(x+3)(x+4) = x2+7x+12
x2 - 7x = 60 x2 - 7x - 60 = 0 (x - 12) (x + 5) = 0 x = 12 or x = -5
x2 + 7x = 2x2 + 6 x2 - 2x2 + 7x = 6 -x2 +7x - 6 = 0 or alternatively x2 - 7x + 6 = 0
x2 + x2 = 2x2
x2+7x+12=0 x2+7x=-12 x+Sqaure Root of 7x= 2(SQ Root Symbol)3
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
6x3 - x2 + 17 = 2x2 + 47 6x3 - x2 - 2x2 = 47 - 17 x2(6x - 1 - 2) = 30 this is the simplest factorisation.
x2 + 7x + 9 = 3 ∴ x2 + 7x + 6 = 0 ∴ (x + 6)(x + 1) = 0 ∴ x ∈ {-6, -1}
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)
3
x2+7x+12
x2+7x+3 = 0 Using the quadratic equation formula the solutions are:- x = -6.541381265 or x = -0.4586187349
x2 + x2 = 2x2
x2+11x+11 = 7x+9 x2+11x-7x+11-9 = 0 x2+4x+2 = 0 The above quadratic equation can be solved by using the quadratic equation formula and it will have two solutions.