Mixing
Even-even-odd rule with multiple terms in the radicand
$begingroup$
I found one definition of the even-even-odd rule by the peeps at planetmath.
They say that if
- a real variable to an even exponent is under a radical
- and the radical has an even index
- and, when the radical is eliminated, the resulting exponent on the variable is odd
then absolute value signs must be placed around the variable.
My question is how to apply the rule when the radicand and result after eliminating the radical have multiple terms.
Do you wrap the whole radicand with absolute value signs or how does that work?
Here's the problem that provoked this question:
Rationalize the expression:
$sqrt{x^2 + x} - sqrt{x^2 - x}$
radicals
asked Jun 8 '17 at 17:12
Brad TurekBrad Turek
1034
$endgroup$
add a comment |
$begingroup$
I found one definition of the even-even-odd rule by the peeps at planetmath.
They say that if
- a real variable to an even exponent is under a radical
- and the radical has an even index
- and, when the radical is eliminated, the resulting exponent on the variable is odd
then absolute value signs must be placed around the variable.
My question is how to apply the rule when the radicand and result after eliminating the radical have multiple terms.
Do you wrap the whole radicand with absolute value signs or how does that work?
Here's the problem that provoked this question:
Rationalize the expression:
$sqrt{x^2 + x} - sqrt{x^2 - x}$
radicals
asked Jun 8 '17 at 17:12
Brad TurekBrad Turek
1034
$endgroup$
add a comment |
$begingroup$
I found one definition of the even-even-odd rule by the peeps at planetmath.
They say that if
- a real variable to an even exponent is under a radical
- and the radical has an even index
- and, when the radical is eliminated, the resulting exponent on the variable is odd
then absolute value signs must be placed around the variable.
My question is how to apply the rule when the radicand and result after eliminating the radical have multiple terms.
Do you wrap the whole radicand with absolute value signs or how does that work?
Here's the problem that provoked this question:
Rationalize the expression:
$sqrt{x^2 + x} - sqrt{x^2 - x}$
radicals
asked Jun 8 '17 at 17:12
Brad TurekBrad Turek
1034
$endgroup$
I found one definition of the even-even-odd rule by the peeps at planetmath.
They say that if
- a real variable to an even exponent is under a radical
- and the radical has an even index
- and, when the radical is eliminated, the resulting exponent on the variable is odd
then absolute value signs must be placed around the variable.
My question is how to apply the rule when the radicand and result after eliminating the radical have multiple terms.
Do you wrap the whole radicand with absolute value signs or how does that work?
Here's the problem that provoked this question:
Rationalize the expression:
$sqrt{x^2 + x} - sqrt{x^2 - x}$
radicals
radicals
asked Jun 8 '17 at 17:12
Brad TurekBrad Turek
1034
asked Jun 8 '17 at 17:12
Brad TurekBrad Turek
1034
edited Jun 8 '17 at 17:22
Brad Turek
asked Jun 8 '17 at 17:12
Brad TurekBrad Turek
1034
asked Jun 8 '17 at 17:12
Brad TurekBrad Turek
1034
asked Jun 8 '17 at 17:12
Brad TurekBrad Turek
1034
1034
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
you can multiply $$sqrt{x^2+x}-sqrt{x^2-x}$$ by $$frac{sqrt{x^2+x}+sqrt{x^2+x}}{sqrt{x^2+x}+sqrt{x^2+x}}$$
answered Jun 8 '17 at 17:16
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.9k42867
$endgroup$
$begingroup$
Thanks, I did get that far. What I want to know is, once I multiply by that conjugate, how to go about applying the even-even-odd rule to something with multiple terms.
$endgroup$
– Brad Turek
Jun 8 '17 at 17:19
add a comment |
$begingroup$
The exponent in question is the exponent on the whole expression under the radical. The point is that in $sqrt{a^2}$ we are guaranteed that $a^2 ge 0$ so taking the square root makes sense. As $sqrt {a^2}$ is defined to be the positive square root, we need an absolute value sign on the $a$. $sqrt{a^2}=a$ is incorrect for $a le 0$.
In your case, if we are just given
$$sqrt{(x^2+x)^2}-sqrt{(x^2-x)^2}$$
we need to put absolute value signs around the expressions because $x^2+x, x^2-x$ might be less than zero, so we get
$$|x^2+x|-|x^2-x|$$
However if we multiply $$(sqrt{x^2+x}-sqrt{x^2-x})(sqrt{x^2+x}+sqrt{x^2-x})=sqrt{(x^2+x)^2}-sqrt{(x^2-x)^2}$$
the original problem tells us that $x^2+x ge 0, x^2-x ge 0$ or the original expression does not make sense. We can use that to avoid the absolute value signs and get $$(x^2+x)-(x^2-x)=2x$$
Note that for $x=frac 12$,
$$sqrt{(x^2+x)^2}-sqrt{(x^2-x)^2}=sqrt{(frac 34)^2}-sqrt{(frac {-1}4)^2}=|frac 34|-|frac {-1}4|=frac 12 neq 2x$$
but here we had nothing to tell us $x^2-x ge 0$
answered Jun 8 '17 at 17:44
Ross MillikanRoss Millikan
301k24200375
$endgroup$
$begingroup$
Thanks for the great answer! Maybe it's totally obvious, but could you elaborate on how the original problem "tells us that $x^2+x ge 0, x^2-x ge 0$"
$endgroup$
– Brad Turek
Jun 8 '17 at 18:03
$begingroup$
They are both provided to you under the square root sign, so they can't be negative.
$endgroup$
– Ross Millikan
Jun 8 '17 at 18:28
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
you can multiply $$sqrt{x^2+x}-sqrt{x^2-x}$$ by $$frac{sqrt{x^2+x}+sqrt{x^2+x}}{sqrt{x^2+x}+sqrt{x^2+x}}$$
answered Jun 8 '17 at 17:16
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.9k42867
$endgroup$
$begingroup$
Thanks, I did get that far. What I want to know is, once I multiply by that conjugate, how to go about applying the even-even-odd rule to something with multiple terms.
$endgroup$
– Brad Turek
Jun 8 '17 at 17:19
add a comment |
$begingroup$
you can multiply $$sqrt{x^2+x}-sqrt{x^2-x}$$ by $$frac{sqrt{x^2+x}+sqrt{x^2+x}}{sqrt{x^2+x}+sqrt{x^2+x}}$$
answered Jun 8 '17 at 17:16
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.9k42867
$endgroup$
$begingroup$
Thanks, I did get that far. What I want to know is, once I multiply by that conjugate, how to go about applying the even-even-odd rule to something with multiple terms.
$endgroup$
– Brad Turek
Jun 8 '17 at 17:19
add a comment |
$begingroup$
you can multiply $$sqrt{x^2+x}-sqrt{x^2-x}$$ by $$frac{sqrt{x^2+x}+sqrt{x^2+x}}{sqrt{x^2+x}+sqrt{x^2+x}}$$
answered Jun 8 '17 at 17:16
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.9k42867
$endgroup$
you can multiply $$sqrt{x^2+x}-sqrt{x^2-x}$$ by $$frac{sqrt{x^2+x}+sqrt{x^2+x}}{sqrt{x^2+x}+sqrt{x^2+x}}$$
answered Jun 8 '17 at 17:16
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.9k42867
answered Jun 8 '17 at 17:16
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.9k42867
answered Jun 8 '17 at 17:16
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.9k42867
answered Jun 8 '17 at 17:16
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
78.9k42867
78.9k42867
$begingroup$
Thanks, I did get that far. What I want to know is, once I multiply by that conjugate, how to go about applying the even-even-odd rule to something with multiple terms.
$endgroup$
– Brad Turek
Jun 8 '17 at 17:19
add a comment |
$begingroup$
Thanks, I did get that far. What I want to know is, once I multiply by that conjugate, how to go about applying the even-even-odd rule to something with multiple terms.
$endgroup$
– Brad Turek
Jun 8 '17 at 17:19
$begingroup$
Thanks, I did get that far. What I want to know is, once I multiply by that conjugate, how to go about applying the even-even-odd rule to something with multiple terms.
$endgroup$
– Brad Turek
Jun 8 '17 at 17:19
$begingroup$
Thanks, I did get that far. What I want to know is, once I multiply by that conjugate, how to go about applying the even-even-odd rule to something with multiple terms.
$endgroup$
– Brad Turek
Jun 8 '17 at 17:19
add a comment |
$begingroup$
The exponent in question is the exponent on the whole expression under the radical. The point is that in $sqrt{a^2}$ we are guaranteed that $a^2 ge 0$ so taking the square root makes sense. As $sqrt {a^2}$ is defined to be the positive square root, we need an absolute value sign on the $a$. $sqrt{a^2}=a$ is incorrect for $a le 0$.
In your case, if we are just given
$$sqrt{(x^2+x)^2}-sqrt{(x^2-x)^2}$$
we need to put absolute value signs around the expressions because $x^2+x, x^2-x$ might be less than zero, so we get
$$|x^2+x|-|x^2-x|$$
However if we multiply $$(sqrt{x^2+x}-sqrt{x^2-x})(sqrt{x^2+x}+sqrt{x^2-x})=sqrt{(x^2+x)^2}-sqrt{(x^2-x)^2}$$
the original problem tells us that $x^2+x ge 0, x^2-x ge 0$ or the original expression does not make sense. We can use that to avoid the absolute value signs and get $$(x^2+x)-(x^2-x)=2x$$
Note that for $x=frac 12$,
$$sqrt{(x^2+x)^2}-sqrt{(x^2-x)^2}=sqrt{(frac 34)^2}-sqrt{(frac {-1}4)^2}=|frac 34|-|frac {-1}4|=frac 12 neq 2x$$
but here we had nothing to tell us $x^2-x ge 0$
answered Jun 8 '17 at 17:44
Ross MillikanRoss Millikan
301k24200375
$endgroup$
$begingroup$
Thanks for the great answer! Maybe it's totally obvious, but could you elaborate on how the original problem "tells us that $x^2+x ge 0, x^2-x ge 0$"
$endgroup$
– Brad Turek
Jun 8 '17 at 18:03
$begingroup$
They are both provided to you under the square root sign, so they can't be negative.
$endgroup$
– Ross Millikan
Jun 8 '17 at 18:28
add a comment |
$begingroup$
The exponent in question is the exponent on the whole expression under the radical. The point is that in $sqrt{a^2}$ we are guaranteed that $a^2 ge 0$ so taking the square root makes sense. As $sqrt {a^2}$ is defined to be the positive square root, we need an absolute value sign on the $a$. $sqrt{a^2}=a$ is incorrect for $a le 0$.
In your case, if we are just given
$$sqrt{(x^2+x)^2}-sqrt{(x^2-x)^2}$$
we need to put absolute value signs around the expressions because $x^2+x, x^2-x$ might be less than zero, so we get
$$|x^2+x|-|x^2-x|$$
However if we multiply $$(sqrt{x^2+x}-sqrt{x^2-x})(sqrt{x^2+x}+sqrt{x^2-x})=sqrt{(x^2+x)^2}-sqrt{(x^2-x)^2}$$
the original problem tells us that $x^2+x ge 0, x^2-x ge 0$ or the original expression does not make sense. We can use that to avoid the absolute value signs and get $$(x^2+x)-(x^2-x)=2x$$
Note that for $x=frac 12$,
$$sqrt{(x^2+x)^2}-sqrt{(x^2-x)^2}=sqrt{(frac 34)^2}-sqrt{(frac {-1}4)^2}=|frac 34|-|frac {-1}4|=frac 12 neq 2x$$
but here we had nothing to tell us $x^2-x ge 0$
answered Jun 8 '17 at 17:44
Ross MillikanRoss Millikan
301k24200375
$endgroup$
$begingroup$
Thanks for the great answer! Maybe it's totally obvious, but could you elaborate on how the original problem "tells us that $x^2+x ge 0, x^2-x ge 0$"
$endgroup$
– Brad Turek
Jun 8 '17 at 18:03
$begingroup$
They are both provided to you under the square root sign, so they can't be negative.
$endgroup$
– Ross Millikan
Jun 8 '17 at 18:28
add a comment |
$begingroup$
The exponent in question is the exponent on the whole expression under the radical. The point is that in $sqrt{a^2}$ we are guaranteed that $a^2 ge 0$ so taking the square root makes sense. As $sqrt {a^2}$ is defined to be the positive square root, we need an absolute value sign on the $a$. $sqrt{a^2}=a$ is incorrect for $a le 0$.
In your case, if we are just given
$$sqrt{(x^2+x)^2}-sqrt{(x^2-x)^2}$$
we need to put absolute value signs around the expressions because $x^2+x, x^2-x$ might be less than zero, so we get
$$|x^2+x|-|x^2-x|$$
However if we multiply $$(sqrt{x^2+x}-sqrt{x^2-x})(sqrt{x^2+x}+sqrt{x^2-x})=sqrt{(x^2+x)^2}-sqrt{(x^2-x)^2}$$
the original problem tells us that $x^2+x ge 0, x^2-x ge 0$ or the original expression does not make sense. We can use that to avoid the absolute value signs and get $$(x^2+x)-(x^2-x)=2x$$
Note that for $x=frac 12$,
$$sqrt{(x^2+x)^2}-sqrt{(x^2-x)^2}=sqrt{(frac 34)^2}-sqrt{(frac {-1}4)^2}=|frac 34|-|frac {-1}4|=frac 12 neq 2x$$
but here we had nothing to tell us $x^2-x ge 0$
answered Jun 8 '17 at 17:44
Ross MillikanRoss Millikan
301k24200375
$endgroup$
The exponent in question is the exponent on the whole expression under the radical. The point is that in $sqrt{a^2}$ we are guaranteed that $a^2 ge 0$ so taking the square root makes sense. As $sqrt {a^2}$ is defined to be the positive square root, we need an absolute value sign on the $a$. $sqrt{a^2}=a$ is incorrect for $a le 0$.
In your case, if we are just given
$$sqrt{(x^2+x)^2}-sqrt{(x^2-x)^2}$$
we need to put absolute value signs around the expressions because $x^2+x, x^2-x$ might be less than zero, so we get
$$|x^2+x|-|x^2-x|$$
However if we multiply $$(sqrt{x^2+x}-sqrt{x^2-x})(sqrt{x^2+x}+sqrt{x^2-x})=sqrt{(x^2+x)^2}-sqrt{(x^2-x)^2}$$
the original problem tells us that $x^2+x ge 0, x^2-x ge 0$ or the original expression does not make sense. We can use that to avoid the absolute value signs and get $$(x^2+x)-(x^2-x)=2x$$
Note that for $x=frac 12$,
$$sqrt{(x^2+x)^2}-sqrt{(x^2-x)^2}=sqrt{(frac 34)^2}-sqrt{(frac {-1}4)^2}=|frac 34|-|frac {-1}4|=frac 12 neq 2x$$
but here we had nothing to tell us $x^2-x ge 0$
answered Jun 8 '17 at 17:44
Ross MillikanRoss Millikan
301k24200375
answered Jun 8 '17 at 17:44
Ross MillikanRoss Millikan
301k24200375
answered Jun 8 '17 at 17:44
Ross MillikanRoss Millikan
301k24200375
answered Jun 8 '17 at 17:44
Ross MillikanRoss Millikan
301k24200375
301k24200375
$begingroup$
Thanks for the great answer! Maybe it's totally obvious, but could you elaborate on how the original problem "tells us that $x^2+x ge 0, x^2-x ge 0$"
$endgroup$
– Brad Turek
Jun 8 '17 at 18:03
$begingroup$
They are both provided to you under the square root sign, so they can't be negative.
$endgroup$
– Ross Millikan
Jun 8 '17 at 18:28
add a comment |
$begingroup$
Thanks for the great answer! Maybe it's totally obvious, but could you elaborate on how the original problem "tells us that $x^2+x ge 0, x^2-x ge 0$"
$endgroup$
– Brad Turek
Jun 8 '17 at 18:03
$begingroup$
They are both provided to you under the square root sign, so they can't be negative.
$endgroup$
– Ross Millikan
Jun 8 '17 at 18:28
$begingroup$
Thanks for the great answer! Maybe it's totally obvious, but could you elaborate on how the original problem "tells us that $x^2+x ge 0, x^2-x ge 0$"
$endgroup$
– Brad Turek
Jun 8 '17 at 18:03
$begingroup$
Thanks for the great answer! Maybe it's totally obvious, but could you elaborate on how the original problem "tells us that $x^2+x ge 0, x^2-x ge 0$"
$endgroup$
– Brad Turek
Jun 8 '17 at 18:03
$begingroup$
They are both provided to you under the square root sign, so they can't be negative.
$endgroup$
– Ross Millikan
Jun 8 '17 at 18:28
$begingroup$
They are both provided to you under the square root sign, so they can't be negative.
$endgroup$
– Ross Millikan
Jun 8 '17 at 18:28
add a comment |
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