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Find zeros of the function.. check my answers? f(x)= x^2+16x+62?

pita0001

New member
Feb 9, 2014
18
Find the zeros of the following function:

\(\displaystyle f(x)= x^2+16x+62\)

I got -6, -10 using completing the square.
 

Jameson

Administrator
Staff member
Jan 26, 2012
4,041
I got -6, -10 using completing the square
Hmm, let's see if that works.

\(\displaystyle (x+6)(x+10)=x^2+10x+6x+60=x^2+16x+60\)

That's not the same as the problem you posted. Is there a typo maybe?
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
I edited your original post to include the problem in the body of the post. We ask that this is done for the sake of clarity. When I first read your post, I thought to myself, "where is the function?" :D
 

pita0001

New member
Feb 9, 2014
18
The function is: f(x)= x^2+16x+62

I used completing the square method and my answers were -6 and -10.
 

Jameson

Administrator
Staff member
Jan 26, 2012
4,041
That's not correct then. Can you show us your work?
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
The function is: f(x)= x^2+16x+62

I used completing the square method and my answers were -6 and -10.
As Jameson demonstrated, those cannot be the zeroes of the given function. The rational roots theorem says that for this function, if there are rational roots, they must be factors of 62, and neither of -6 nor -10 are such factors.

Can you post your work so we can figure out where you went astray?
 

pita0001

New member
Feb 9, 2014
18
-16+- (sq root) 16^2-4(1)62)/2(1)

-16+- (sq root) 256-248/2

-16+- (sq root) 8/2




-16+8/2

=-4



-16-8/2
= -12

- - - Updated - - -

Or would my final answer be


-8+rad(2)?
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
It looks to me like you are using the quadratic formula, as I see no evidence of completing the square. If we use the quadratic formula, then we identify:

\(\displaystyle a=1,\,b=16,\,c=62\)

and we have:

\(\displaystyle x=\frac{-16\pm\sqrt{16^2-4\cdot1\cdot62}}{2\cdot1}\)

Simplify:

\(\displaystyle x=\frac{-16\pm\sqrt{8}}{2}\)

Now, can you simplify further?

It looks like you are almost there...although you have omitted the $\pm$ sign. I really couldn't be sure what you were doing until I did it myself since you did not include bracketing symbols which leaves room for ambiguity. :D
 

pita0001

New member
Feb 9, 2014
18
Can divide by 2, so,

-8+- 2(sq root) 2


right?
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Well, we have:

\(\displaystyle x=\frac{-16\pm2\sqrt{2}}{2}\)

Now, we divide by 2...what do we have?