x2 + 12x - 11 = x2 + 12x + (12/2)2 - (12/2)2 - 11 = (x + 6)2 - 47
36.1. You take the coefficient of x : = 122. Halve it : = 63. Then square it : = 364. Add it.This gives x2 + 12x + 36 = (x + 6)2The above method only works if the coefficient of x2 is 1. If it is not then the processs is slightly more complicated.
-2x3 + 2x2 + 12x =(-2x) (x2 - x - 6) =(-2x) (x+2) (x-3)
I assume you mean,X2 + 12X - 5 = 0X2 + 12X = 5Now, halve the linear term (12), square it and add it to both sidesX2 + 12X + 36 = 5 + 36factor on the left, gather terms on the right(X + 6)2 = 41(X + 6)2 - 41 = 0==============vertex form.... (- 6, - 41)========================--------------vertexAnd that number added is 36
if x2 + 7 = 37, then x2 = 29 and x = ±√29
x2+12x+36=0 2x+12x+36=0 14x+36=0 14x=-36 x=-36/14 x=-2 4/7
x2 - 12x + 36 = (x - 6)2
x2+12x+28Improved Answer:--x2+12x+28 = (-x+14)(x+2) when factored
x=5/2*sqr(3)
x2+12x+5 = 3 x2+12x+5-3 = 0 x2+12x+2 = 0
x2 + 12x + 27 = (x + 9)(x + 3)
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
=>2x - 12x + 36 = 0 =>10x = 36=>x = 3.6Another contributor's answer:This is a quadratic equation question:x2-12x+36 = 0When factorised:(x-6)(x-6) = 0
x2 + 12x + 32 = (x + 8)(x + 4)
x2 + 12x + 32 = (x + 4)(x + 8)
3x3+12x = 3x(x2+4)
x2-12x+35 = (x-5)(x-7)