2016年5月23日 · where #x_{root1}# comes from using the #pm# as subtraction and #x_{root2}# comes from using the #pm# as addition. 2) Factor, for simple equations with #a=1# , for equations with simple integer roots we can find the factors by looking for a two numbers with add to #b# and multiply to #c# (there is a modification to these method used for ...

2016年7月31日 · Put on a common denominator: (2(x^2 - x - 6))/(2x^2) =(x(x - 6))/(2x^2) + (2x(2x + 12))/(2x^2) 2x^2 - 2x - 12 = x^2 - 6x + 4x^2 + 24x 0 = 3x^2 + 20x + 12 0 = 3x^2 ...

2017年7月2日 · Moving the #-6# to the right side: #x^2+x=6# Completing the square by adding #(b/2)^2# to both sides, where #b# is the second coefficient (in this case, #b=1#): #x^2+x+(1/2)^2=6+(1/2)^2# #x^2+x+1/4=6+1/4# #(x+1/2)^2=25/4# Simplifying: #x+1/2=\pm5/2# #x=-1/2\pm5/2# #x=4/2# or #-6/2# #x=2# or #-3#

The vertex is (1/2-25/4) or (1/2, -6 1/4). y=x^2-x-6 is a quadratic equation in the form of ax^2+b^2+6, where a=1, b=-1, and c=-6. The vertex is the minimum or maximum point of the equation. The x value can be found using the formula x=(-b)/(2a). The y value can be found by substituting the value for x into the equation and solving for y.

2017年4月14日 · x^4 Let's write the equation in this form (x^4*x^2)/(x^2) Now numerator's (x^2) will cancel the one in the ...

2017年2月11日 · x=-2, 8 1) Rearrange into the form" "ax^2+bx+c=0 2) Factorise 3) Put each bracket =0 and solve each bracket 1) Rearrange x^2-16=6x subtract 6x from both sides x^2-16-6x=6x-6x x^2-16-6x=0 x^2-6x-16=0 2) Factorise x^2-6x-16=0 a) "because it is an " x^2 " equation, the brackets will be of the form " (x+-" ")(x+-" ") we now need to find the two numbers for the brackets. x^2color(red)(-6)xcolor ...

2016年6月1日 · How do you find the excluded values of #\frac{2x+1}{x^2-x-6}#? What are Excluded Values for Rational Expressions? How do you state the excluded values for rational expressions?

2016年9月7日 · (x-2)/(x+6) The first step is to factorise both numerator and denominator. Both are quadratics and can be ...

2016年3月15日 · (1/2, -13/2) Vertex of a parabola in the form ax^2+bx+c is given by: x=-b/(2a) Note this is only gives the x-coordinate; we will have to evaluate this value to get the y-coordinate.

2016年7月12日 · x = 6 and x = -2 By moving 4x to the left of our equation, we can then complete the square in the following way: x^2 = 4x + 12 x^2 -4x = 12 We take the coefficient on the x-term, namely -4, divide it by 2, and square the result, giving us (-4/2)^(2) = (-2)^2 = 4 Since our goal is to rewrite our equation in the form of x^2 -4x + ? = 12 + …