Why doesn't x^2+1 has no zeroes

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Discussion Overview

The discussion revolves around the polynomial equation ##x^2 + 1## and the question of why it does not have any real zeroes. Participants explore the properties of the polynomial, including its behavior for both positive and negative inputs, and touch on related concepts such as the discriminant and the nature of roots in quadratic equations.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • Some participants argue that ##x^2 + 1## is always positive, as it evaluates to at least 1 for all real numbers, thus it cannot be zero.
  • Others point out that the equation ##x^2 + 1 = 0## leads to ##x^2 = -1##, indicating that the solutions are imaginary numbers (##x = i## or ##x = -i##), which are not real.
  • There is a discussion about the polynomial ##x^2 - 1##, with some participants asserting it has two distinct real roots, while others challenge the method used to demonstrate this.
  • Some participants mention the discriminant formula ##D = b^2 - 4ac## and its implications for determining the nature of roots in quadratic equations.
  • One participant expresses confusion about the concepts discussed, indicating a lack of familiarity with certain mathematical topics, such as the discriminant and factoring polynomials.
  • Participants clarify that when the discriminant is zero, it indicates two equal roots rather than just one root.

Areas of Agreement / Disagreement

Participants generally agree that ##x^2 + 1## has no real zeroes, but there is disagreement regarding the explanation and understanding of the roots of ##x^2 - 1##, as well as the implications of the discriminant.

Contextual Notes

Some participants express uncertainty about the mathematical concepts discussed, indicating varying levels of familiarity with quadratic equations and their properties.

Who May Find This Useful

This discussion may be useful for individuals interested in understanding the nature of polynomial equations, the concept of real versus imaginary roots, and the application of the discriminant in determining the nature of roots in quadratic equations.

manogyana25
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Why doesn't x^2+1 has no zeroes i.e, zero of the polynomial??
 
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##x^2+1## is always positive. For a negative number -k, you get
##f(-k)=(-k)^2+1=k^2+1##
For positive numbers also, it will always return a positive number.
Since ##x^2>=0##
##x^2+1>=1##
Hence it can never be zero.

{This part is optional:
If you know calculus,
##\frac{dy}{dx}=2x##
For critical points,##2x=0## so x=0 is a point of inflection
##d^2y/dx^2=2## which is positive. So at x=0, you have a minimum value.
##f(0)=1##
So the least possible value attained by the polynomial is 1. }

Hence, the curve never goes below the y=1.So the polynomial can never be zero.

Take this as a question for practice:
Prove that ##x^2-1## has two distinct zeroes
 
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Technically, f(x) = x2+1 has no real zeroes. :-p
 
AdityaDev said:
##x^2+1## is always positive. For a negative number -k, you get
##f(-k)=(-k)^2+1=k^2+1##
For positive numbers also, it will always return a positive number.
Since ##x^2>=0##
##x^2+1>=1##
Hence it can never be zero.

{This part is optional:
If you know calculus,
##\frac{dy}{dx}=2x##
For critical points,##2x=0## so x=0 is a point of inflection
##d^2y/dx^2=2## which is positive. So at x=0, you have a minimum value.
##f(0)=1##
So the least possible value attained by the polynomial is 1. }

Hence, the curve never goes below the y=1.So the polynomial can never be zero.

Take this as a question for practice:
Prove that ##x^2-1## has two distinct zeroes
X^2-1=0
X^2=1
X=√1
X=1
Hey, I have another thought regarding this..
X^2+1=0
X^2=-1
X=√-1 which is an imaginary number..
So it can never be zero..
 
Last edited:
manogyana25 said:
Hey, I have another thought regarding this..
X^2+1=0
X^2=-1
X=√-1 which is an imaginary number..
So it can never be zero..
Correct.

The method with which you tried to show that x^2-1 has two real distinct zeroes is wrong.
x^2 can also be zero. But it does not have distinct roots.
For distinct roots, the curve has to come below the x-axis. That is, the parabola should have a negative minimum.
Or you can use this: b^2 - 4ac>0
 
AdityaDev said:
Correct.

The method with which you tried to show that x^2-1 has two real distinct zeroes is wrong.
x^2 can also be zero. But it does not have distinct roots.
For distinct roots, the curve has to come below the x-axis. That is, the parabola should have a negative minimum.
Or you can use this: b^2 - 4ac>0
Sorry.. Whatever the curve, parabola and everything you said, I didn't understand. I haven't took those classes yet. Can you please name that chapter for me??
 
Ok.
Do you know this formula:
$$D=b^2-4ac$$
It is called discriminant.
When D<0, you have no real roots.
When D>0, you have 2 distinct roots.
When D=0, you have 1 root.
 
What is D
 
manogyana25 said:
What is D
Discriminant.
Do you know how to find roots of a quadratic equation using this formula:
$$x=\frac{-b^+_-\sqrt{b^2-4ac}}{2a}$$
 
  • #10
Noooo... I've not taken classes regarding that topic yet.. If you really want to say that to me.. Then probably you must spend all your time here... ;)
 
  • #11
K
 
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  • #12
manogyana25 said:
X^2-1=0
X^2=1
X=√1
X=1
This is not correct (or at least, not complete).
If x2 = 1,
then x2 - 1 = 0
so (x - 1)(x + 1) = 0
so x = 1 or x = -1
The point is, there are two solutions.
manogyana25 said:
Hey, I have another thought regarding this..
X^2+1=0
X^2=-1
X=√-1 which is an imaginary number..
So it can never be zero..
Similar to the above, if x2 + 1 = 0, then x = i or x = -i. Again, there are two solutions, neither of which is real.
 
Last edited:
  • #13
Oh yes, I totally missed this point... I got it now.. Thank you Mark44.. :)
 
  • #14
Mark44 said:
This is not correct (or at least, not complete).
If x2 = 1,
then x2 - 1 = 0
so (x - 1)(x + 1) = 0
so x = 1 or x = -1
The point is, there are two solutions.

Similar to the above, if x2 + 1 = 0, then x = i or x = -i. Again, there are two solutions, neither of which is real.
Hey but its not (x^2-1)^2 ... Its x^2 -1 right?? Then how can it be (x+1)(x-1)
 
Last edited by a moderator:
  • #15
manogyana25 said:
Hey but its not (x^2-1)^2 ... Its x^2 -1 right?? Then how can it be (x+1)(x-1)
x2 - 1 factors into (x + 1)(x - 1). To verify this for yourself, carry out the multiplication of (x + 1)(x - 1).
 
  • #16
Yes yes I got it thanks!
 
  • #17
When D is zero, it doesn't mean there's only one root
It means there are two roots but both of them are equal!UchihaClan13
 

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