2013年6月1日 · 3. You might not understand the parts of a fraction nor the notation: 2/x=12. The 2 is the numerator, and tells the number of parts of the whole that you have. The line is the, …

2015年2月3日 · We can use the Sum-Product method. Bring everything to one side: x^2-x=12->x^2-x-12=0 We now have a quadratic equation of the form ax^2+bx+c=0 where a=1, b=-1 …

2015年6月19日 · It is not necessary, but later steps are easier if we first make sure that we have 0 on one side. (Any constant will be ok, but 0 is easiest.) Partition the number line by finding …

2015年11月26日 · The you need to get one #x# on its own on one side of = and everything else on the other side. This is how you do it for your question: When multiplying, change what you …

2015年10月25日 · x=1 First, rearrange the equation like this: 2^(2x) + 2^(x + 2) - 12 = 0 (2^x)^2 + 4(2^x) - 12 = 0 Now, treat this like a quadratic equation by substituting 2^x=s: s^2+4s-12=0 …

2016年2月6日 · x^2-x-12 = (x-4)(x+3) Look for a pair of factors of 12 that differ by 1. The pair 4, 3 works. Hence we find: x^2-x-12 = (x-4)(x+3) In general: (x-a)(x+b) = x^2-(a-b)x-ab In our case …

2016年3月9日 · #x^2+x-12=0# This is in the form of a Quadratic equation (in form #ax^2+bx+c=0#) Where. #a=1,b=1,c=-12 ...

2017年2月8日 · x^2-x-12>0" "=>" "x<"-"3 uu x>4. Since the right-hand side (RHS) is already 0, we start by factoring the left-hand side (LHS): " "x^2-x-12>0 =>(x-4)(x+3)>0 In its factored form, …

2017年12月23日 · x=-3" or "x=4 >"given the graph of "x^2-x-12 "the solutions are the values of x where the graph crosses" "the x-axis" graph{(y-x^2+x+12)((x+3)^2+(y-0)^2-0.04)((x-4)^2+(y …

2016年8月15日 · The first step is to take out a #color(blue)"common factor"# of 2. #rArr2(x^2-x-6)=0# We now require to factorise the #color(blue)"quadratic"# in the bracket. Find the factors …