Question

HELP :)
Determine the x intercepts of the quadratic function f(x)= x^2+12x+37

Answer 1

Looking at this equation, we can notice something very troublesome with trying to factor this guy. 37 is a prime number, only divisible by itself and 1.
In order to solve this guy, you will need to use a technique called completing the square. After that, the function should be easier to factor, then you can determine the x intercepts.

Answer 2

CAN YOU PLEASE HELP ME? :)

Answer 3

complete the square i mean :) i suck at it lol

Answer 4

http://www.purplemath.com/modules/sqrquad.htm
^this shows a good method for using the method I described.

Because we want to find the x intercepts, we set y(x)=0 to get
0=x^2+12x+37

First take 37 and put it on the other side of the equation to get:
-37=x^2+12x

Then, divide by the coefficient in front of the x^2 term to get
-37=x^2+12x -I know this is the same eq as above, it is just part of the process

Now, take half of the coefficient in front of the x term and square it
12/2=6 6^2=36 (be careful not to forget the original sign of this term!!)

Add the term from ^ to each side of the original equation to get
-37+36= x^2+12x+36

Factor:
-1=(x+6)^2

Normally, here we would take the square root and solve for x, however we cannot take the square root of a negative number. Thus, this function has no x intercepts.

If you have access to a graphing calculator or software, try plugging in this function and graphing it. You will find that this graph does not intersect the x axis at any point

Answer 5

This one is tricky because the question implies that there ARE x intercepts, however this cannot be true due to the computations above

Answer 6

So there no intercepts? :)

Answer 7

hey just use quadratic formula :)

Answer 8

Yes, no intercepts.

@some_someone, if you tried to do so with this equation you would find yourself trying to take the square root of a negative, just like we found ourselves doing in the above computations.

Answer 9

exactly, you get complex/imaginary solutions. There is no actual real solutions :D

Answer 10

lol, I don't think that was the focus of this exercise. However we basically learned HOW to complete the square and found something interesting about this function. What math are you taking @HopelessMathStudent ?

Answer 11

btw, the complex solution would be
(+/-)i-6=x, just in case you need to know the imaginary solution

Answer 12

yep :)

Answer 13

You wouldn't know that if you didn't complete the square ;) lol

Answer 14

haha i did the quadratic formula method :P

Answer 15

Both methods are useful :D just depends on what you are comfortable with and what will lead to the correct(and sensible) solution

Answer 16

agreed