x2-3x+2=0 [x^2-2X(3/2)X(x)]+2=0 Notice that the first two terms have been brought under the structure a^2 -2ab. So you need b^2 to complete the square and here b^2-(3/2)^2 = 9/4 Therefore [x^2-2X(3/2)X(x)+(9/4-9/4)]+2=0 [x^2-2X(3/2)X(x)+(9/4]+(-9/4+2)=0 (x-3/2)^2 = 9/4-2 = 1/4 Taking the sqroot (x-3/2)= +1/2 or -1/2 (x-3/2)= +1/2 gives x = 1/2+3/2 = 2 (x-3/2)= -1/2 gives x = -1/2+3/2 = 1 Answer: x = 2 and x=1 solve by using the quadratic equasion x2-3x+2=0 x2+[(-2x)+(-x)]+2=0 [splitting the middle term into two parts whose multiplication is the product of the square term and the constant term. -3x = (-2x) + (-x) so that (-2x)X(-x) = +(2x^2) = (x^2)X(2)] x2-2x-x+2=0 (x2-2x)-x+2=0 x(x-2)-(x-2)=0 xp-p=0 where p = (x-2) p(x-1) = 0 (x-2)(x-1) = 0 (x-2) = 0 gives x = 2 and (x-1) = 0 gives x = 1
Check all your bulbs mate, I bet there is one blown thats why your fogs are on, or you have a faulty fog light switch. What do you mean about the breaks, 'they grip'?