Hello,
I am doing a practice test for school. I ran across the equation (3xsquared -18 +24 =0) in which it says to solve. I started to go about it like a quadratic equation and find 2 numbers that add up to -18 and product =24 but can't! Am I going about this the wrong way? Is their another way to solve this equation?
Please help
That is incorrect. You would seek "two numbers that add to -18 and product=24" only if the equation were of the form \(\displaystyle x^2-18x+ 24\)- that is, only if the coefficient of \(\displaystyle x^2\) were equal to 1. More generally, if you have a quadratic \(\displaystyle ax^2+ bx+ c\) and want to factor as \(\displaystyle (px+ q)(rx+ s)= prx^2+ (qr+ps)x+ qs\), you would have to find four number, p, q, r, and s, such that pr= a, qr+ps= b, qs= c. Here, with a= 3, b= -18 (you don't have the "x" but that clearly what you mean), and c= 24, you would want to find p, q, r, and s such that pr= 3, qr+ ps= -18, and qs= 24. In order to do this with integers (which might NOT be possible but there are just too many alternatives if we allow fractions irrational numbers) then, with pr= 3, either p= 3 and r= 1 or p= 1 and r= 3. With p= 3, r= 1,qr+ ps=q+ 3s= -18 and qs= 24. Looking at the various factorings of 24 (-24*-1, -12*-2, -8*-3, -6*-4) (we know q and s have the same sign because qs is positive and we know they must be negative because q+ 3s is negative) we see that
q= -6, s= -4. That is, \(\displaystyle 3x^2- 18x+ 24= (3x- 6)(x- 4)\)
As I said, not all (actually very few) can be factored with integer coefficients. In that case, we can use "completing the square" or the "quadratic formula". If you do not know those, you will probably learn them soon.
