Prove x² + x + 1 > 0 with Complete the Square

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Homework Help Overview

The discussion revolves around proving that the expression x² + x + 1 is greater than zero for all values of x, utilizing the method of completing the square. Participants explore the implications of the quadratic's behavior and its minimum value.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the method of completing the square and its relevance to proving the expression is always positive. Questions arise regarding the existence of real zeros and the implications of complex numbers. Some participants suggest focusing on the minimum point of the quadratic and its implications for the overall sign of the expression.

Discussion Status

There is a mix of interpretations regarding the approach to the problem. Some participants have provided insights into the nature of the quadratic function and its minimum value, while others question the necessity of discussing whether the function is increasing or decreasing. The conversation reflects a productive exploration of the problem without reaching a definitive conclusion.

Contextual Notes

Participants note that the problem may involve assumptions about the nature of the quadratic function and its zeros, which are under discussion. The original poster expresses uncertainty about how to proceed with the proof.

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Homework Statement



Show x² + x + 1 > 0
well this is basically asking to prove that any value of x you put in will give a result bigger than zero, but how do i prove it

Homework Equations


According to the markscheme you should use complete the sqare


The Attempt at a Solution



(X + 1/2) + 3/4

but how does that help

Thanks :)
 
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Can you determine whether this has any real zeros? What if it doesn't?
 
err don't see what you mean. there should be nothing to do with complex numbers in this if that's where your heading?

Thanks :)
 
thomas49th said:

Homework Statement



Show x² + x + 1 > 0
well this is basically asking to prove that any value of x you put in will give a result bigger than zero, but how do i prove it

Homework Equations


According to the markscheme you should use complete the sqare


The Attempt at a Solution



(X + 1/2) + 3/4

but how does that help

Thanks :)

(X + 1/2)2 + 3/4

Meaning that the graph's min pt is 3/4. Therefore it is >0.
 
qazxsw11111 said:
(X + 1/2)2 + 3/4

Meaning that the graph's min pt is 3/4. Therefore it is >0.
You'll also have to show that the function is increasing.
 
to show it's increasing you'll have to have somthing > 3/4 right?


(X + 1/2)² + 3/4 = 0
x + 1/2 = sqrt(-3/4)

you can't root a minus :o

Is that what you have to do?
 
The solution has nothing to do with showing this is increasing - it can't, because [tex]x^2 + x + 1[/tex] is not an increasing function of [tex]x[/tex].


There are only a few possibilities for a quadratic:

1. It does not have any real zeros - in this case it is always positive or always negative

2. It has one real zero - in this case it is either positive and zero in one spot, or it is negative and zero in one spot

3. It has exactly two real zeros - in this case it is positive for some values of x, negative for some, and zero for 2.

So, if you can determine how many real zeros this has, and if you can determine at its sign at a value of [tex]x[/tex], you are done. Do not waste your time worrying about whether it is increasing or decreasing - that is the wrong track.
 
thomas49th said:

Homework Statement



Show x² + x + 1 > 0
well this is basically asking to prove that any value of x you put in will give a result bigger than zero, but how do i prove it

Homework Equations


According to the markscheme you should use complete the sqare


The Attempt at a Solution



(X + 1/2) + 3/4

but how does that help

Thanks :)

You've shown x^2+x+1=(x+1/2)^2+3/4. (x+1/2)^2 is nonnegative (it's a SQUARE). 3/4 is positive. Hence the sum is positive.
 
qazxsw11111 said:
(X + 1/2)2 + 3/4

Meaning that the graph's min pt is 3/4. Therefore it is >0.

Yes, absolutely. At this point, the problem is solved. The squared quantity can be zero at minimum, therefore the whole expression has a minimum of 3/4. In other words, it's always positive.

End of discussion?
 
  • #10
yes thankyou very much :)
 

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