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, "vinculum," and it means, "divided by." The, "x," is the denominator. It tells how many parts are in the whole object. 3. So, look at the equation again. Solve: 2/x=12 4.

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 and c=-12 We find two numbers that will give c=-12 as a product and b=-1 as a sum (or difference). We can try 1*12, 2*6,3*4 3and4 will fit, with a -sign to the 4, as that would make the sum 3-4= …

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 zeros of the expression and points at which it is not defined. x^2+x-12 is defined for all x so we only need to find its zeros Solve x^2+x-12=0 by factoring if possible.

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 want to get rid of into 1 as anything multiplied by 1 does not change. #color(blue)("Method for getting rid of the 2 on the left.")# Divide both sides by 2

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 (s+6)(s-2)=0 s=-6 or s=2 Now, go back to the substitution: 2^x=s 2^x=-6 or 2^x=2 Since 2^x can never equal a negative number, we can rule out the first solution. However, the second …

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 we found a=4, b=3 giving a-b=1 and ab=12

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, this inequality tells us that the product of two numbers (x-4) and (x+3) is positive (greater than 0). In order for a product of two terms to be positive, either both terms must be positive or both terms …

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-0)^2-0.04)=0 [-10, 10, -5, 5]} rArrx=-3" or "x=4" are the solutions" "we can verify this algebraically by solving" x^2-x-12=0 rArr(x-4)(x+3)=0 x-4=0rArrx=4 x+3=0rArrx=-3

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 of - 6 which sum to - 1. These are - 3 and +2. #rArr2x^2-2x-12=2(x-3)(x+2)=0# Either of the factors (x - 3) , (x +2) can equal zero. solve : #x-3=0rArrx=3# solve : #x+2 ...