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

[manogyana25]

Why doesn't x^2+1 has no zeroes i.e, zero of the polynomial??

 

[AdityaDev]

$x^2+1$ 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

 

[da_nang]

Technically, f(x) = x²+1 has no real zeroes.

 

[manogyana25]

$x^2+1$ 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..

 

[AdityaDev]

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

 

[manogyana25]

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??

 

[AdityaDev]

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.

 

[manogyana25]

What is D

 

[AdityaDev]

What is D

Discriminant.
Do you know how to find roots of a quadratic equation using this formula:
$$x=\frac{-b^+_-\sqrt{b^2-4ac}}{2a}$$

 

[manogyana25]

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... ;)

 

[manogyana25]

K

 

[Mark44]

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.

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.

 

[manogyana25]

Oh yes, I totally missed this point... I got it now.. Thank you Mark44.. :)

 

[manogyana25]

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)

 

[Mark44]

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).

 

[manogyana25]

Yes yes I got it thanks!

 

[UchihaClan13]

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