Calculate the roots of -t^2-2t+1

  • Thread starter DottZakapa
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In summary: Incorrect, the negative sign in front of the quadratic term is lost when the expression is set equal to zero.The incorrect answer is the second or the first.
  • #1
DottZakapa
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Homework Statement
-t^2-2t+1
Relevant Equations
roots are
I have to compute the roots in order to compute an integral partial fraction decomposition

##\frac {2 \pm 2 \sqrt {4+4}} {-2} = -1 \mp \sqrt 2##

the correct on is

## (t+1- \sqrt 2)(+1+ \sqrt 2) \\or \\
(t+1+ \sqrt 2)(+1- \sqrt 2)##

the general rule is
 
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  • #2
Are you trying to factorise that quadratic expression?
 
  • #3
PeroK said:
Are you trying to factorise that quadratic expression?
yes
 
  • #4
DottZakapa said:
yes

The answer to the either/or is neither. There need to be two linear factors involving ##t##. Such as ##(t - r_1)(t-r_2)## where ##r_1, r_2## are the roots.

If you think you have a factorisation, you can multiply it out to see whether you are correct.
 
  • #5
##\frac {2 \pm 2 \sqrt {2}} {-2} = -1 \mp \sqrt 2##
this is correct right?
 
  • #6
DottZakapa said:
##\frac {2 \pm 2 \sqrt {4+4}} {-2} = -1 \mp \sqrt 2##
this is correct right?

No. It's arithmetically wrong, I'm sorry to say.
 
  • #7
sorry
PeroK said:
No. It's arithmetically wrong, I'm sorry to say.
sorry i made a typing mistake, I've corrected the rest too
this is what i meant
##\frac {2 \pm 2 \sqrt {2}} {-2} = -1 \mp \sqrt 2##
 
  • #8
DottZakapa said:
sorry

sorry i made a typing mistake, I've corrected the rest too
this is what i meant
##\frac {2 \pm 2 \sqrt {2}} {-2} = -1 \mp \sqrt 2##

These are now the roots of your quadratic.
 
  • #9
PeroK said:
These are now the roots of your quadratic.
so between the two the correct is the second or the first
## (t+1- \sqrt 2)(+1+ \sqrt 2) \\or \\
(t+1+ \sqrt 2)(+1- \sqrt 2)##
that minus in front of r1 multiplies both signs (the one in front of 1 and square root ) or just the one in front of one? that's my question
 
  • #10
DottZakapa said:
so between the two the correct is the second or the first
## (t+1- \sqrt 2)(+1+ \sqrt 2) \\or \\
(t+1+ \sqrt 2)(+1- \sqrt 2)##
that minus in front of r1 multiplies both signs (the one in front of 1 and square root ) or just the one in front of one? that's my question
If you put a ##t## in the second term, where it should be, then there is no difference.

## (t+1- \sqrt 2)(t+1+ \sqrt 2) \\or \\
(t+1+ \sqrt 2)(t+1- \sqrt 2)##

These are equal.
 
  • #11
PeroK said:
If you put a ##t## in the second term, where it should be, then there is no difference.

## (t+1- \sqrt 2)(t+1+ \sqrt 2) \\or \\
(t+1+ \sqrt 2)(t+1- \sqrt 2)##

These are equal.
so if i multiply both or just the first doesn't make a difference
 
  • #12
DottZakapa said:
so if i multiply both or just the first doesn't make a difference
Those two expressions are equal because it doesn't matter what order you multiply the factors. In general:

##(t-a)(t-b) = (t-b)(t-a)##

And, in general, for expressions involving numbers:

##XY = YX##

where ##X## and ##Y## can be any numbers or expressions involving numbers.

This property of numbers is called "commuting". It works for both addition and multiplication of numbers, as we also have:

##X + Y = Y + X##.
 
  • #13
PeroK said:
If you put a ##t## in the second term, where it should be, then there is no difference.

## (t+1- \sqrt 2)(t+1+ \sqrt 2) \\or \\
(t+1+ \sqrt 2)(t+1- \sqrt 2)##

These are equal.
They may be equal, but neither is correct because they don't give the expected result when multiplied out. Note that the original expression is -t^2-2t+1. There is a negative sign in front of the quadratic term that is lost when the expresion is set equal to zero in order to find the roots. This doesn't affect the roots but it changes the overall sign in front of the desired partial fractions.
 
  • #14
kuruman said:
They may be equal, but neither is correct because they don't give the expected result when multiplied out. Note that the original expression is -t^2-2t+1. There is a negative sign in front of the quadratic term that is lost when the expresion is set equal to zero in order to find the roots. This doesn't affect the roots but it changes the overall sign in front of the desired partial fractions.
could you please tell me exactly where the mistake is, I'm getting so confused with this equation
 
  • #15
DottZakapa said:
could you please tell me exactly where the mistake is, I'm getting so confused with this equation

##-t^2 -2t +1 = - (t+1- \sqrt 2)(t+1+ \sqrt 2)##
 

1. What is the equation being used to calculate the roots of -t^2-2t+1?

The equation being used is the quadratic formula, which is (-b±√(b^2-4ac))/2a, where a, b, and c are the coefficients of the quadratic equation.

2. How do I know if the roots of -t^2-2t+1 are real or complex?

You can determine if the roots are real or complex by calculating the discriminant, which is b^2-4ac. If the discriminant is a positive number, the roots will be real. If it is a negative number, the roots will be complex.

3. Can the roots of -t^2-2t+1 be irrational numbers?

Yes, the roots can be irrational numbers. This means that the roots cannot be expressed as a fraction of two integers and will have an infinite number of decimal places.

4. What if the coefficient of t^2 is 0?

If the coefficient of t^2 is 0, the equation will become a linear equation and the roots can be found by solving for t. There will only be one root in this case.

5. Is there a specific order to follow when using the quadratic formula to calculate the roots of -t^2-2t+1?

Yes, the order of operations when using the quadratic formula is to first find the values of a, b, and c from the equation. Then, plug these values into the formula and follow the order of operations (PEMDAS) to calculate the roots.

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