GREhelp wrote:
Sorry for the inconvenience, the source of the question is
Manhattan Prep GRE Algebra Practice book and the answer was 6. It was a fill in the blank question.
I understand how you solved it, the explanation makes perfect sense. Where I struggled was that after substituting the 4 into the problem, and getting -11. I did not know what to do next. How did you know that you need to set the equation to -11?
The first thing I tried to do when I first saw this problem was to try to factor it out but how do you factor a number that is adding +13. The only two numbers that multiply to equal 13 is (1 and 13) however they do not sum to 10.
THANKS SO MUCH!!
Please Grehelp: formatting properly the question: title, questions, and answer choices.
As for the question, you do need to find your unknown that is K. As such, substituting x=4 in your equation it becomes \(4^2\)-10*4+13=K >>> K=-11
Then you do have \(x^2\)-10X+13=-11 >>>>>> \(x^2\) -10X+13+11=0 >>>> \(x^2\)-10X+24=0
The best and faster way to solve for
Hope this helps