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Title: Solve this quadratic equation(X^2+14x+33=0)? Post by: darkyoda1589 on Sep 9, 2013 X^2+14x+33=0
I solved it as ss={11,3} but on mathway (http://mathway.com/problem.aspx?p=precalculus) it says {-11,-3} which one is right? Oh,I know know what went wrong. I solved by quadratic equation, and I began it with a b,I simply forgot the (-). :") Thanks all. Title: Solve this quadratic equation(X^2+14x+33=0)? Post by: robbaron on Sep 9, 2013 x^2+14x+33
(x+11)(x+3)=0 x+11=0; x+3=0 x=-11; x=-3 The latter. Title: Solve this quadratic equation(X^2+14x+33=0)? Post by: deep3636 on Sep 9, 2013 Content hidden
Title: Solve this quadratic equation(X^2+14x+33=0)? Post by: rmalviya on Sep 9, 2013 I think where you went wrong is when you had it (correctly) factored as
(x+11)(x+3) = 0 ..you forgot to change the signs... If we multiply 2 numbers and get 0, then at least one of those numbers must be 0, so x+11 = 0 or x+3 = 0 Solving these for x, we get x = -11 or -3 Title: Solve this quadratic equation(X^2+14x+33=0)? Post by: socalkid1 on Sep 9, 2013 x^2+14x+33=0
Factor the "c" term which is 33. We get: 1*33 and 3*11 We will use 3*11 because that is the factor pair that adds up to the "b" term, which is 14. (3+11=14) So this equation factors into: (x+3)(x+11)=0 Using the zero product rule, we can split this into two equations: x+3=0 and x+11=0 Solving for x in both equations, you get the two solutions to be: x=-3 and x=-11 Title: Solve this quadratic equation(X^2+14x+33=0)? Post by: robbcfd on Sep 9, 2013 One hint that may help you to check your work in the future is to look at the signs in the equation. Since the second sign is positive, you know both signs in the factored equation must be the same. Add to this information that the first sign is positive and you know that when you factor the equation, both signs will be positive (and thus both the solutions must be negative). If the first sign of the original equation had been negative and the second positive, both signs in the factored equation would be negative and the solutions both positive. However if the second sign in the original equation were negative, the factored solution would have one positive and one negative sign with a positive and negative solution. This would have shown you that there was a problem with your ss={11,3} . The information can be very helpful when taking timed tests. You can quickly eliminate wrong answers and may not need to spend time working the whole solution.
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