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

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Discussion Overview

The discussion revolves around finding the zeros of the quadratic function f(x) = x^2 + 16x + 62. Participants explore various methods for solving the equation, particularly focusing on completing the square and the quadratic formula.

Discussion Character

  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant claims to have found the zeros as -6 and -10 using the completing the square method.
  • Another participant questions the validity of these zeros, noting that they do not match the original function when factored.
  • A third participant points out that the rational roots theorem indicates that if there are rational roots, they must be factors of 62, which -6 and -10 are not.
  • One participant provides a detailed breakdown of using the quadratic formula, suggesting that the zeros can be derived from the expression x = (-16 ± √(16² - 4(1)(62))) / (2(1)).
  • Another participant attempts to simplify the quadratic formula results but omits the ± sign, leading to further clarification requests from others.
  • There is a discussion about simplifying the expression further, with participants working through the algebraic steps together.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the correct zeros of the function. Disagreement exists regarding the initial answers provided and the methods used to solve the equation.

Contextual Notes

There are unresolved mathematical steps and ambiguities in the expressions presented, particularly regarding the use of the quadratic formula and the simplification process.

pita0001
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Find the zeros of the following function:

$$f(x)= x^2+16x+62$$

I got -6, -10 using completing the square.
 
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pita0001 said:
I got -6, -10 using completing the square

Hmm, let's see if that works.

$$(x+6)(x+10)=x^2+10x+6x+60=x^2+16x+60$$

That's not the same as the problem you posted. Is there a typo maybe?
 
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
 
The function is: f(x)= x^2+16x+62

I used completing the square method and my answers were -6 and -10.
 
That's not correct then. Can you show us your work?
 
pita0001 said:
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?
 
-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)?
 
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:

$$a=1,\,b=16,\,c=62$$

and we have:

$$x=\frac{-16\pm\sqrt{16^2-4\cdot1\cdot62}}{2\cdot1}$$

Simplify:

$$x=\frac{-16\pm\sqrt{8}}{2}$$

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
 
Can divide by 2, so,

-8+- 2(sq root) 2right?
 
  • #10
Well, we have:

$$x=\frac{-16\pm2\sqrt{2}}{2}$$

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

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