Question 63379: Solve for x by completing the square:
kx^2-3cx-4a=0
I have tried but I wonder if there is no solution. Thanks for your time!
By the way, this question is not from any textbook. My teacher had it on the board for the class to solve.
Answer by ptaylor(2198)   (Show Source):
You can put this solution on YOUR website! We will solve for x by completing the square:
kx^2-3cx-4a=0 First, divide both sides by k(assuming k is nonzero). We get:
x^2-(3c/k)x-4(a/k)=0 Next, add 4(a/k) to both sides and we have:
x^2-(3c/k)x=4(a/k) Now we add (9c^2/4k^2) to both sides. Why? Because we want to make the left side of the equation a perfect square. (9c^2/4k^2)was selected because the sum of the factors (-3c/2k)^2 will yield the second term.
x^2-(3c/k)x+(9c^2/4k^2)=4(a/k)+(9c^2/4k^2) Now we factor the left side and get:
(x-3c/2k)^2=4(a/k)+(9c^2/4k^2) Now, we take the sqrt of both sides and we have
x-3c/2k=+or-sqrt(4(a/k)+(9c^2/4k^2)) simplifying the left side and adding (3c/2k) to both sides, we get
x=3c/2k+or-sqrt((16ak+9c^2)/4k^2) further simplifying, we have
x=(3c+or-sqrt(16ak+9c^2))/2k
You can re-check the work but this is my approach to solving quadratics by completing the square.
When factoring quadratics, it useful to remember that when the first term is one, the sum of the factors of the third term will equal the second term.
Example: Find the factors of x^2-6x+9. By inspection, we see that 3^2 is 9 and 3+3is 6 so the factors are (x-3)(x-3)
Hope this helps and have a nice holiday season---ptaylor
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