Mixing
Simplifying Radicals In Heron’s Formula
$begingroup$
When I sometimes use Heron’s formula and the Pythagorean Theorem in the coordinate system to find the area of a triangle, I get stumped at the last step: simplifying the radical. Is there a general way to do it?
Edit:
An example will be to find the area of a triangle with side lengths $sqrt{2},sqrt{3},sqrt{5}$, which is simplifying${{sqrt{(sqrt{2}+sqrt{3}+sqrt{5})(-sqrt{2}+sqrt{3}+sqrt{5})(sqrt{2}-sqrt{3}+sqrt{5})(sqrt{2}+sqrt{3}-sqrt{5})}}over 4}$.
analytic-geometry radicals
asked Jan 27 at 0:02
Math LoverMath Lover
17210
$endgroup$
add a comment |
$begingroup$
When I sometimes use Heron’s formula and the Pythagorean Theorem in the coordinate system to find the area of a triangle, I get stumped at the last step: simplifying the radical. Is there a general way to do it?
Edit:
An example will be to find the area of a triangle with side lengths $sqrt{2},sqrt{3},sqrt{5}$, which is simplifying${{sqrt{(sqrt{2}+sqrt{3}+sqrt{5})(-sqrt{2}+sqrt{3}+sqrt{5})(sqrt{2}-sqrt{3}+sqrt{5})(sqrt{2}+sqrt{3}-sqrt{5})}}over 4}$.
analytic-geometry radicals
asked Jan 27 at 0:02
Math LoverMath Lover
17210
$endgroup$
$begingroup$
Please show some examples.
$endgroup$
– marty cohen
Jan 27 at 0:10
$begingroup$
A search of Math.SE for "simplify radical" yields many results. Perhaps some of them could be useful to you.
$endgroup$
– Blue
Jan 27 at 0:18
add a comment |
$begingroup$
When I sometimes use Heron’s formula and the Pythagorean Theorem in the coordinate system to find the area of a triangle, I get stumped at the last step: simplifying the radical. Is there a general way to do it?
Edit:
An example will be to find the area of a triangle with side lengths $sqrt{2},sqrt{3},sqrt{5}$, which is simplifying${{sqrt{(sqrt{2}+sqrt{3}+sqrt{5})(-sqrt{2}+sqrt{3}+sqrt{5})(sqrt{2}-sqrt{3}+sqrt{5})(sqrt{2}+sqrt{3}-sqrt{5})}}over 4}$.
analytic-geometry radicals
asked Jan 27 at 0:02
Math LoverMath Lover
17210
$endgroup$
When I sometimes use Heron’s formula and the Pythagorean Theorem in the coordinate system to find the area of a triangle, I get stumped at the last step: simplifying the radical. Is there a general way to do it?
Edit:
An example will be to find the area of a triangle with side lengths $sqrt{2},sqrt{3},sqrt{5}$, which is simplifying${{sqrt{(sqrt{2}+sqrt{3}+sqrt{5})(-sqrt{2}+sqrt{3}+sqrt{5})(sqrt{2}-sqrt{3}+sqrt{5})(sqrt{2}+sqrt{3}-sqrt{5})}}over 4}$.
analytic-geometry radicals
analytic-geometry radicals
asked Jan 27 at 0:02
Math LoverMath Lover
17210
asked Jan 27 at 0:02
Math LoverMath Lover
17210
edited Jan 27 at 1:46
Math Lover
asked Jan 27 at 0:02
Math LoverMath Lover
17210
asked Jan 27 at 0:02
Math LoverMath Lover
17210
asked Jan 27 at 0:02
Math LoverMath Lover
17210
17210
$begingroup$
Please show some examples.
$endgroup$
– marty cohen
Jan 27 at 0:10
$begingroup$
A search of Math.SE for "simplify radical" yields many results. Perhaps some of them could be useful to you.
$endgroup$
– Blue
Jan 27 at 0:18
add a comment |
$begingroup$
Please show some examples.
$endgroup$
– marty cohen
Jan 27 at 0:10
$begingroup$
A search of Math.SE for "simplify radical" yields many results. Perhaps some of them could be useful to you.
$endgroup$
– Blue
Jan 27 at 0:18
$begingroup$
Please show some examples.
$endgroup$
– marty cohen
Jan 27 at 0:10
$begingroup$
Please show some examples.
$endgroup$
– marty cohen
Jan 27 at 0:10
$begingroup$
A search of Math.SE for "simplify radical" yields many results. Perhaps some of them could be useful to you.
$endgroup$
– Blue
Jan 27 at 0:18
$begingroup$
A search of Math.SE for "simplify radical" yields many results. Perhaps some of them could be useful to you.
$endgroup$
– Blue
Jan 27 at 0:18
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Hint:
See if you can show
$$(sqrt{a}+sqrt{b}+sqrt{c})(-sqrt{a}+sqrt{b}+sqrt{c})(sqrt{a}-sqrt{b}+sqrt{c})(sqrt{a}+sqrt{b}-sqrt{c})=-a^2-b^2-c^2+2ab+2ac+2bc$$
Edit:
Let $$F(x,y,z)=(x+y+z)(-x+y+z)(x-y+z)(x+y-z).$$ We will show that $$F(x,y,z)=-x^4-y^4-z^4+2x^2y^2+2x^2z^2+2y^2z^2.$$ At first I wanted to use calculus to compare coefficients using derivatives but it is easier than this.
Note that $F(-x,y,z)=F(x,y,z)$ and remains unchanged by any permutation of $x,y,z$. This means $F(x,y,z)$ contains only even powers of $x,y,z$. Since every term is of degree $4$, the only possible summands are of the form $kx^4$ (and therefore by symmetry $ky^4$ and $kz^4$) or $ell x^2y^2$ (and therefore $ell x^2z^2$ and $ell y^2z^2$.)
So $$F(x,y,z)=k(x^4+y^4+z^4)+ell(x^2y^2+x^2z^2+y^2z^2).$$
To calculate $k$, just plug in $y=z=0$ to find $k=-1$. To calculate $ell$, plug in $x=y=z=1$. Then on the one hand $F(1,1,1)=(1+1+1)(-1+1+1)^3=3$, and on the other $$F(1,1,1)= -(1^4+1^4+1^4)+ell(1^2 1^2 +1^2 1^2+1^2 1^2)=-3+3ell$$
Solving for $ell$, we get $ell=2$.
edited Jan 30 at 15:00
Math Lover
17210
answered Jan 27 at 16:47
Cheerful ParsnipCheerful Parsnip
21.1k23598
$endgroup$
$begingroup$
Whoa! Why is that?
$endgroup$
– Math Lover
Jan 27 at 19:44
$begingroup$
There's probably a sexy proof, but you can just laboriously multiply everything out.
$endgroup$
– Cheerful Parsnip
Jan 27 at 19:50
1
$begingroup$
I suspected there would be a lot of cancellation so I had Wolfram Alpha multiply everything out for me. wolframalpha.com/input/…
$endgroup$
– Cheerful Parsnip
Jan 27 at 20:03
1
$begingroup$
@MathLover, it turns out the proof is actually very simple. I edited the answer to provide more details.
$endgroup$
– Cheerful Parsnip
Jan 29 at 23:52
1
$begingroup$
Whoa, that is so cool! Thanks!
$endgroup$
– Math Lover
Jan 30 at 14:12
|
show 5 more comments
$begingroup$
To simplify a radical expression, assuming that the radicand is a positive real number, find the radicand's largest perfect square factor and rewrite the expression as a product of the two factors. Simplify the perfect square that you factored out and you are left with the radical in simplest radical form.
I think the best way to see this is by example.
Consider $sqrt{50}$.
Since $sqrt{ab}=sqrt{a}sqrt{b}$, this can be rewritten as $sqrt{25}sqrt{2}$. Note that $25$ is the largest perfect square factor of $50$. Now you can easily take the square root of $25$, which is $pm{5}$. Since our original expression is positive, our new expression in simplest radical form is ${5}sqrt{2}$.
answered Jan 27 at 0:10
GnumbertesterGnumbertester
670114
$endgroup$
1
$begingroup$
Good answer, although I'll quibble on your last line. There are two numbers whose square is 50, one positive and one negative, but the square root of 50 is the nonnegative one. So $sqrt{50} = 5sqrt{2}$, not $pm5sqrt{2}$.
$endgroup$
– Matthew Leingang
Jan 27 at 0:13
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint:
See if you can show
$$(sqrt{a}+sqrt{b}+sqrt{c})(-sqrt{a}+sqrt{b}+sqrt{c})(sqrt{a}-sqrt{b}+sqrt{c})(sqrt{a}+sqrt{b}-sqrt{c})=-a^2-b^2-c^2+2ab+2ac+2bc$$
Edit:
Let $$F(x,y,z)=(x+y+z)(-x+y+z)(x-y+z)(x+y-z).$$ We will show that $$F(x,y,z)=-x^4-y^4-z^4+2x^2y^2+2x^2z^2+2y^2z^2.$$ At first I wanted to use calculus to compare coefficients using derivatives but it is easier than this.
Note that $F(-x,y,z)=F(x,y,z)$ and remains unchanged by any permutation of $x,y,z$. This means $F(x,y,z)$ contains only even powers of $x,y,z$. Since every term is of degree $4$, the only possible summands are of the form $kx^4$ (and therefore by symmetry $ky^4$ and $kz^4$) or $ell x^2y^2$ (and therefore $ell x^2z^2$ and $ell y^2z^2$.)
So $$F(x,y,z)=k(x^4+y^4+z^4)+ell(x^2y^2+x^2z^2+y^2z^2).$$
To calculate $k$, just plug in $y=z=0$ to find $k=-1$. To calculate $ell$, plug in $x=y=z=1$. Then on the one hand $F(1,1,1)=(1+1+1)(-1+1+1)^3=3$, and on the other $$F(1,1,1)= -(1^4+1^4+1^4)+ell(1^2 1^2 +1^2 1^2+1^2 1^2)=-3+3ell$$
Solving for $ell$, we get $ell=2$.
edited Jan 30 at 15:00
Math Lover
17210
answered Jan 27 at 16:47
Cheerful ParsnipCheerful Parsnip
21.1k23598
$endgroup$
$begingroup$
Whoa! Why is that?
$endgroup$
– Math Lover
Jan 27 at 19:44
$begingroup$
There's probably a sexy proof, but you can just laboriously multiply everything out.
$endgroup$
– Cheerful Parsnip
Jan 27 at 19:50
1
$begingroup$
I suspected there would be a lot of cancellation so I had Wolfram Alpha multiply everything out for me. wolframalpha.com/input/…
$endgroup$
– Cheerful Parsnip
Jan 27 at 20:03
1
$begingroup$
@MathLover, it turns out the proof is actually very simple. I edited the answer to provide more details.
$endgroup$
– Cheerful Parsnip
Jan 29 at 23:52
1
$begingroup$
Whoa, that is so cool! Thanks!
$endgroup$
– Math Lover
Jan 30 at 14:12
|
show 5 more comments
$begingroup$
Hint:
See if you can show
$$(sqrt{a}+sqrt{b}+sqrt{c})(-sqrt{a}+sqrt{b}+sqrt{c})(sqrt{a}-sqrt{b}+sqrt{c})(sqrt{a}+sqrt{b}-sqrt{c})=-a^2-b^2-c^2+2ab+2ac+2bc$$
Edit:
Let $$F(x,y,z)=(x+y+z)(-x+y+z)(x-y+z)(x+y-z).$$ We will show that $$F(x,y,z)=-x^4-y^4-z^4+2x^2y^2+2x^2z^2+2y^2z^2.$$ At first I wanted to use calculus to compare coefficients using derivatives but it is easier than this.
Note that $F(-x,y,z)=F(x,y,z)$ and remains unchanged by any permutation of $x,y,z$. This means $F(x,y,z)$ contains only even powers of $x,y,z$. Since every term is of degree $4$, the only possible summands are of the form $kx^4$ (and therefore by symmetry $ky^4$ and $kz^4$) or $ell x^2y^2$ (and therefore $ell x^2z^2$ and $ell y^2z^2$.)
So $$F(x,y,z)=k(x^4+y^4+z^4)+ell(x^2y^2+x^2z^2+y^2z^2).$$
To calculate $k$, just plug in $y=z=0$ to find $k=-1$. To calculate $ell$, plug in $x=y=z=1$. Then on the one hand $F(1,1,1)=(1+1+1)(-1+1+1)^3=3$, and on the other $$F(1,1,1)= -(1^4+1^4+1^4)+ell(1^2 1^2 +1^2 1^2+1^2 1^2)=-3+3ell$$
Solving for $ell$, we get $ell=2$.
edited Jan 30 at 15:00
Math Lover
17210
answered Jan 27 at 16:47
Cheerful ParsnipCheerful Parsnip
21.1k23598
$endgroup$
$begingroup$
Whoa! Why is that?
$endgroup$
– Math Lover
Jan 27 at 19:44
$begingroup$
There's probably a sexy proof, but you can just laboriously multiply everything out.
$endgroup$
– Cheerful Parsnip
Jan 27 at 19:50
1
$begingroup$
I suspected there would be a lot of cancellation so I had Wolfram Alpha multiply everything out for me. wolframalpha.com/input/…
$endgroup$
– Cheerful Parsnip
Jan 27 at 20:03
1
$begingroup$
@MathLover, it turns out the proof is actually very simple. I edited the answer to provide more details.
$endgroup$
– Cheerful Parsnip
Jan 29 at 23:52
1
$begingroup$
Whoa, that is so cool! Thanks!
$endgroup$
– Math Lover
Jan 30 at 14:12
|
show 5 more comments
$begingroup$
Hint:
See if you can show
$$(sqrt{a}+sqrt{b}+sqrt{c})(-sqrt{a}+sqrt{b}+sqrt{c})(sqrt{a}-sqrt{b}+sqrt{c})(sqrt{a}+sqrt{b}-sqrt{c})=-a^2-b^2-c^2+2ab+2ac+2bc$$
Edit:
Let $$F(x,y,z)=(x+y+z)(-x+y+z)(x-y+z)(x+y-z).$$ We will show that $$F(x,y,z)=-x^4-y^4-z^4+2x^2y^2+2x^2z^2+2y^2z^2.$$ At first I wanted to use calculus to compare coefficients using derivatives but it is easier than this.
Note that $F(-x,y,z)=F(x,y,z)$ and remains unchanged by any permutation of $x,y,z$. This means $F(x,y,z)$ contains only even powers of $x,y,z$. Since every term is of degree $4$, the only possible summands are of the form $kx^4$ (and therefore by symmetry $ky^4$ and $kz^4$) or $ell x^2y^2$ (and therefore $ell x^2z^2$ and $ell y^2z^2$.)
So $$F(x,y,z)=k(x^4+y^4+z^4)+ell(x^2y^2+x^2z^2+y^2z^2).$$
To calculate $k$, just plug in $y=z=0$ to find $k=-1$. To calculate $ell$, plug in $x=y=z=1$. Then on the one hand $F(1,1,1)=(1+1+1)(-1+1+1)^3=3$, and on the other $$F(1,1,1)= -(1^4+1^4+1^4)+ell(1^2 1^2 +1^2 1^2+1^2 1^2)=-3+3ell$$
Solving for $ell$, we get $ell=2$.
edited Jan 30 at 15:00
Math Lover
17210
answered Jan 27 at 16:47
Cheerful ParsnipCheerful Parsnip
21.1k23598
$endgroup$
Hint:
See if you can show
$$(sqrt{a}+sqrt{b}+sqrt{c})(-sqrt{a}+sqrt{b}+sqrt{c})(sqrt{a}-sqrt{b}+sqrt{c})(sqrt{a}+sqrt{b}-sqrt{c})=-a^2-b^2-c^2+2ab+2ac+2bc$$
Edit:
Let $$F(x,y,z)=(x+y+z)(-x+y+z)(x-y+z)(x+y-z).$$ We will show that $$F(x,y,z)=-x^4-y^4-z^4+2x^2y^2+2x^2z^2+2y^2z^2.$$ At first I wanted to use calculus to compare coefficients using derivatives but it is easier than this.
Note that $F(-x,y,z)=F(x,y,z)$ and remains unchanged by any permutation of $x,y,z$. This means $F(x,y,z)$ contains only even powers of $x,y,z$. Since every term is of degree $4$, the only possible summands are of the form $kx^4$ (and therefore by symmetry $ky^4$ and $kz^4$) or $ell x^2y^2$ (and therefore $ell x^2z^2$ and $ell y^2z^2$.)
So $$F(x,y,z)=k(x^4+y^4+z^4)+ell(x^2y^2+x^2z^2+y^2z^2).$$
To calculate $k$, just plug in $y=z=0$ to find $k=-1$. To calculate $ell$, plug in $x=y=z=1$. Then on the one hand $F(1,1,1)=(1+1+1)(-1+1+1)^3=3$, and on the other $$F(1,1,1)= -(1^4+1^4+1^4)+ell(1^2 1^2 +1^2 1^2+1^2 1^2)=-3+3ell$$
Solving for $ell$, we get $ell=2$.
edited Jan 30 at 15:00
Math Lover
17210
answered Jan 27 at 16:47
Cheerful ParsnipCheerful Parsnip
21.1k23598
edited Jan 30 at 15:00
Math Lover
17210
edited Jan 30 at 15:00
Math Lover
17210
edited Jan 30 at 15:00
Math Lover
17210
17210
answered Jan 27 at 16:47
Cheerful ParsnipCheerful Parsnip
21.1k23598
answered Jan 27 at 16:47
Cheerful ParsnipCheerful Parsnip
21.1k23598
answered Jan 27 at 16:47
Cheerful ParsnipCheerful Parsnip
21.1k23598
21.1k23598
$begingroup$
Whoa! Why is that?
$endgroup$
– Math Lover
Jan 27 at 19:44
$begingroup$
There's probably a sexy proof, but you can just laboriously multiply everything out.
$endgroup$
– Cheerful Parsnip
Jan 27 at 19:50
1
$begingroup$
I suspected there would be a lot of cancellation so I had Wolfram Alpha multiply everything out for me. wolframalpha.com/input/…
$endgroup$
– Cheerful Parsnip
Jan 27 at 20:03
1
$begingroup$
@MathLover, it turns out the proof is actually very simple. I edited the answer to provide more details.
$endgroup$
– Cheerful Parsnip
Jan 29 at 23:52
1
$begingroup$
Whoa, that is so cool! Thanks!
$endgroup$
– Math Lover
Jan 30 at 14:12
|
show 5 more comments
$begingroup$
Whoa! Why is that?
$endgroup$
– Math Lover
Jan 27 at 19:44
$begingroup$
There's probably a sexy proof, but you can just laboriously multiply everything out.
$endgroup$
– Cheerful Parsnip
Jan 27 at 19:50
1
$begingroup$
I suspected there would be a lot of cancellation so I had Wolfram Alpha multiply everything out for me. wolframalpha.com/input/…
$endgroup$
– Cheerful Parsnip
Jan 27 at 20:03
1
$begingroup$
@MathLover, it turns out the proof is actually very simple. I edited the answer to provide more details.
$endgroup$
– Cheerful Parsnip
Jan 29 at 23:52
1
$begingroup$
Whoa, that is so cool! Thanks!
$endgroup$
– Math Lover
Jan 30 at 14:12
$begingroup$
Whoa! Why is that?
$endgroup$
– Math Lover
Jan 27 at 19:44
$begingroup$
Whoa! Why is that?
$endgroup$
– Math Lover
Jan 27 at 19:44
$begingroup$
There's probably a sexy proof, but you can just laboriously multiply everything out.
$endgroup$
– Cheerful Parsnip
Jan 27 at 19:50
$begingroup$
There's probably a sexy proof, but you can just laboriously multiply everything out.
$endgroup$
– Cheerful Parsnip
Jan 27 at 19:50
1
1
$begingroup$
I suspected there would be a lot of cancellation so I had Wolfram Alpha multiply everything out for me. wolframalpha.com/input/…
$endgroup$
– Cheerful Parsnip
Jan 27 at 20:03
$begingroup$
I suspected there would be a lot of cancellation so I had Wolfram Alpha multiply everything out for me. wolframalpha.com/input/…
$endgroup$
– Cheerful Parsnip
Jan 27 at 20:03
1
1
$begingroup$
@MathLover, it turns out the proof is actually very simple. I edited the answer to provide more details.
$endgroup$
– Cheerful Parsnip
Jan 29 at 23:52
$begingroup$
@MathLover, it turns out the proof is actually very simple. I edited the answer to provide more details.
$endgroup$
– Cheerful Parsnip
Jan 29 at 23:52
1
1
$begingroup$
Whoa, that is so cool! Thanks!
$endgroup$
– Math Lover
Jan 30 at 14:12
$begingroup$
Whoa, that is so cool! Thanks!
$endgroup$
– Math Lover
Jan 30 at 14:12
|
show 5 more comments
$begingroup$
To simplify a radical expression, assuming that the radicand is a positive real number, find the radicand's largest perfect square factor and rewrite the expression as a product of the two factors. Simplify the perfect square that you factored out and you are left with the radical in simplest radical form.
I think the best way to see this is by example.
Consider $sqrt{50}$.
Since $sqrt{ab}=sqrt{a}sqrt{b}$, this can be rewritten as $sqrt{25}sqrt{2}$. Note that $25$ is the largest perfect square factor of $50$. Now you can easily take the square root of $25$, which is $pm{5}$. Since our original expression is positive, our new expression in simplest radical form is ${5}sqrt{2}$.
answered Jan 27 at 0:10
GnumbertesterGnumbertester
670114
$endgroup$
1
$begingroup$
Good answer, although I'll quibble on your last line. There are two numbers whose square is 50, one positive and one negative, but the square root of 50 is the nonnegative one. So $sqrt{50} = 5sqrt{2}$, not $pm5sqrt{2}$.
$endgroup$
– Matthew Leingang
Jan 27 at 0:13
add a comment |
$begingroup$
To simplify a radical expression, assuming that the radicand is a positive real number, find the radicand's largest perfect square factor and rewrite the expression as a product of the two factors. Simplify the perfect square that you factored out and you are left with the radical in simplest radical form.
I think the best way to see this is by example.
Consider $sqrt{50}$.
Since $sqrt{ab}=sqrt{a}sqrt{b}$, this can be rewritten as $sqrt{25}sqrt{2}$. Note that $25$ is the largest perfect square factor of $50$. Now you can easily take the square root of $25$, which is $pm{5}$. Since our original expression is positive, our new expression in simplest radical form is ${5}sqrt{2}$.
answered Jan 27 at 0:10
GnumbertesterGnumbertester
670114
$endgroup$
1
$begingroup$
Good answer, although I'll quibble on your last line. There are two numbers whose square is 50, one positive and one negative, but the square root of 50 is the nonnegative one. So $sqrt{50} = 5sqrt{2}$, not $pm5sqrt{2}$.
$endgroup$
– Matthew Leingang
Jan 27 at 0:13
add a comment |
$begingroup$
To simplify a radical expression, assuming that the radicand is a positive real number, find the radicand's largest perfect square factor and rewrite the expression as a product of the two factors. Simplify the perfect square that you factored out and you are left with the radical in simplest radical form.
I think the best way to see this is by example.
Consider $sqrt{50}$.
Since $sqrt{ab}=sqrt{a}sqrt{b}$, this can be rewritten as $sqrt{25}sqrt{2}$. Note that $25$ is the largest perfect square factor of $50$. Now you can easily take the square root of $25$, which is $pm{5}$. Since our original expression is positive, our new expression in simplest radical form is ${5}sqrt{2}$.
answered Jan 27 at 0:10
GnumbertesterGnumbertester
670114
$endgroup$
To simplify a radical expression, assuming that the radicand is a positive real number, find the radicand's largest perfect square factor and rewrite the expression as a product of the two factors. Simplify the perfect square that you factored out and you are left with the radical in simplest radical form.
I think the best way to see this is by example.
Consider $sqrt{50}$.
Since $sqrt{ab}=sqrt{a}sqrt{b}$, this can be rewritten as $sqrt{25}sqrt{2}$. Note that $25$ is the largest perfect square factor of $50$. Now you can easily take the square root of $25$, which is $pm{5}$. Since our original expression is positive, our new expression in simplest radical form is ${5}sqrt{2}$.
answered Jan 27 at 0:10
GnumbertesterGnumbertester
670114
edited Jan 27 at 0:15
answered Jan 27 at 0:10
GnumbertesterGnumbertester
670114
answered Jan 27 at 0:10
GnumbertesterGnumbertester
670114
answered Jan 27 at 0:10
GnumbertesterGnumbertester
670114
670114
1
$begingroup$
Good answer, although I'll quibble on your last line. There are two numbers whose square is 50, one positive and one negative, but the square root of 50 is the nonnegative one. So $sqrt{50} = 5sqrt{2}$, not $pm5sqrt{2}$.
$endgroup$
– Matthew Leingang
Jan 27 at 0:13
add a comment |
1
$begingroup$
Good answer, although I'll quibble on your last line. There are two numbers whose square is 50, one positive and one negative, but the square root of 50 is the nonnegative one. So $sqrt{50} = 5sqrt{2}$, not $pm5sqrt{2}$.
$endgroup$
– Matthew Leingang
Jan 27 at 0:13
1
1
$begingroup$
Good answer, although I'll quibble on your last line. There are two numbers whose square is 50, one positive and one negative, but the square root of 50 is the nonnegative one. So $sqrt{50} = 5sqrt{2}$, not $pm5sqrt{2}$.
$endgroup$
– Matthew Leingang
Jan 27 at 0:13
$begingroup$
Good answer, although I'll quibble on your last line. There are two numbers whose square is 50, one positive and one negative, but the square root of 50 is the nonnegative one. So $sqrt{50} = 5sqrt{2}$, not $pm5sqrt{2}$.
$endgroup$
– Matthew Leingang
Jan 27 at 0:13
add a comment |
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$begingroup$
Please show some examples.
$endgroup$
– marty cohen
Jan 27 at 0:10
$begingroup$
A search of Math.SE for "simplify radical" yields many results. Perhaps some of them could be useful to you.
$endgroup$
– Blue
Jan 27 at 0:18