- #1
manogyana25
- 6
- 2
Why doesn't x^2+1 has no zeroes i.e, zero of the polynomial??
Why doesn't x^2+1 has no zeroes
- Thread starter manogyana25
- Start date
In summary, the conversation discusses the fact that ##x^2+1## has no real zeroes, as it is always positive for any value of x. This is proven through the use of calculus and the concept of critical points. The conversation also briefly touches on the topic of finding roots using the quadratic formula and the discriminant. It is also mentioned that the polynomial ##x^2-1## has two distinct zeroes, which is proven using the same methods as above.
- #2
AdityaDev
- 527
- 33
##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
##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
- #3
da_nang
- 137
- 5
Technically, f(x) = x2+1 has no real zeroes. 
- #4
manogyana25
- 6
- 2
X^2-1=0AdityaDev said:
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..
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- #5
AdityaDev
- 527
- 33
Correct.manogyana25 said:
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
- #6
manogyana25
- 6
- 2
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??AdityaDev said:
- #7
AdityaDev
- 527
- 33
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.
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.
- #8
manogyana25
- 6
- 2
What is D
- #9
AdityaDev
- 527
- 33
Discriminant.manogyana25 said:
Do you know how to find roots of a quadratic equation using this formula:
$$x=\frac{-b^+_-\sqrt{b^2-4ac}}{2a}$$
- #10
manogyana25
- 6
- 2
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
manogyana25
- 6
- 2
K
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- #12
Mark44
Mentor
- 37,782
- 10,173
This is not correct (or at least, not complete).manogyana25 said:
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.manogyana25 said:
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- #13
manogyana25
- 6
- 2
Oh yes, I totally missed this point... I got it now.. Thank you Mark44.. :)
- #14
manogyana25
- 6
- 2
Hey but its not (x^2-1)^2 ... Its x^2 -1 right?? Then how can it be (x+1)(x-1)Mark44 said:
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- #15
Mark44
Mentor
- 37,782
- 10,173
x2 - 1 factors into (x + 1)(x - 1). To verify this for yourself, carry out the multiplication of (x + 1)(x - 1).manogyana25 said:
- #16
manogyana25
- 6
- 2
Yes yes I got it thanks!
- #17
UchihaClan13
- 145
- 12
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
It means there are two roots but both of them are equal!UchihaClan13
FAQ: Why doesn't x^2+1 has no zeroes
Why doesn't x^2+1 have any zeroes?
The equation x^2+1 does not have any zeroes because it represents a quadratic function, which always has a minimum value above the x-axis. Therefore, the graph of x^2+1 never crosses the x-axis and does not have any x-intercepts or zeroes.
Can x^2+1 ever have zeroes?
No, x^2+1 can never have zeroes because it is a quadratic function with a minimum value above the x-axis. This means that the graph of x^2+1 never crosses the x-axis and does not have any x-intercepts or zeroes.
What about imaginary or complex solutions?
Even when considering imaginary or complex solutions, x^2+1 still does not have any zeroes. This is because the imaginary or complex solutions would result in points on the y-axis, not the x-axis, which is where the zeroes are located.
What is the discriminant of x^2+1?
The discriminant of x^2+1 is equal to -4, meaning that the quadratic function has no real solutions. This further supports the fact that x^2+1 does not have any zeroes.
Can the coefficient of x^2 be adjusted to make x^2+1 have zeroes?
No, the coefficient of x^2 cannot be adjusted to make x^2+1 have zeroes. The coefficient only affects the shape and position of the parabola, but it does not change the fact that the function has a minimum value above the x-axis and does not cross the x-axis to have any zeroes.
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