Feb 3, 2015 · 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=-1 and …

Jun 1, 2013 · 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.

Feb 6, 2016 · 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

Jun 19, 2015 · 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.

Mar 9, 2016 · #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 ...

Oct 25, 2015 · 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 …

\[{x}^{2}-x=12\] +. > < ...

Nov 26, 2015 · 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

Jan 15, 2016 · #x^2# can only be #x xx x# so we have a starting point of: #color(blue)((x+?)(x+?))# '~~~~~ #color(blue)("Consider the constant of "-12)# This is a product so one of the factors is negative and the other positive. #1xx12=12# #2xx 6=12# #3xx4=12# '~~~~~ #color(blue)("Now consider the "+x". This is "1xx x)# So the difference between the …

Feb 8, 2017 · 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 …