How do I simplify this surd expression?
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How would one convert this problem into an equation that could then be simplified?
A right-angled triangle has shorter side lengths exactly $a^2-b^2$ and $2ab$ units respectively, where a and b are positive real numbers such that a is greater than b. Find an exact expression for the length of the hypotenuse (in appropriate units).
So obviously I would plug those values into the Pythagorean Theorem.
$c = sqrt{(a^2 - b^2)^2 + (2ab)^2}$
So then I try to simplify it but I always have the $sqrt{2}$ left over which isn't in any of the multiple choice questions on MathXL. I have been stuck on this problem for 30 minutes. Any help would be much appreciated.
algebra-precalculus
edited Dec 10 '18 at 18:00
Key Flex
7,84461233
asked Dec 10 '18 at 17:53
Jerry DamsonJerry Damson
111
$endgroup$
add a comment |
$begingroup$
How would one convert this problem into an equation that could then be simplified?
A right-angled triangle has shorter side lengths exactly $a^2-b^2$ and $2ab$ units respectively, where a and b are positive real numbers such that a is greater than b. Find an exact expression for the length of the hypotenuse (in appropriate units).
So obviously I would plug those values into the Pythagorean Theorem.
$c = sqrt{(a^2 - b^2)^2 + (2ab)^2}$
So then I try to simplify it but I always have the $sqrt{2}$ left over which isn't in any of the multiple choice questions on MathXL. I have been stuck on this problem for 30 minutes. Any help would be much appreciated.
algebra-precalculus
edited Dec 10 '18 at 18:00
Key Flex
7,84461233
asked Dec 10 '18 at 17:53
Jerry DamsonJerry Damson
111
$endgroup$
1
$begingroup$
You should add a term $ab$ and make it an absurd expression.
$endgroup$
– T. Bongers
Dec 10 '18 at 17:53
2
$begingroup$
What do you get when you multiply out everything under the radical (surd)? Can you factor it?
$endgroup$
– Doug M
Dec 10 '18 at 17:58
add a comment |
$begingroup$
How would one convert this problem into an equation that could then be simplified?
A right-angled triangle has shorter side lengths exactly $a^2-b^2$ and $2ab$ units respectively, where a and b are positive real numbers such that a is greater than b. Find an exact expression for the length of the hypotenuse (in appropriate units).
So obviously I would plug those values into the Pythagorean Theorem.
$c = sqrt{(a^2 - b^2)^2 + (2ab)^2}$
So then I try to simplify it but I always have the $sqrt{2}$ left over which isn't in any of the multiple choice questions on MathXL. I have been stuck on this problem for 30 minutes. Any help would be much appreciated.
algebra-precalculus
edited Dec 10 '18 at 18:00
Key Flex
7,84461233
asked Dec 10 '18 at 17:53
Jerry DamsonJerry Damson
111
$endgroup$
How would one convert this problem into an equation that could then be simplified?
A right-angled triangle has shorter side lengths exactly $a^2-b^2$ and $2ab$ units respectively, where a and b are positive real numbers such that a is greater than b. Find an exact expression for the length of the hypotenuse (in appropriate units).
So obviously I would plug those values into the Pythagorean Theorem.
$c = sqrt{(a^2 - b^2)^2 + (2ab)^2}$
So then I try to simplify it but I always have the $sqrt{2}$ left over which isn't in any of the multiple choice questions on MathXL. I have been stuck on this problem for 30 minutes. Any help would be much appreciated.
algebra-precalculus
algebra-precalculus
edited Dec 10 '18 at 18:00
Key Flex
7,84461233
asked Dec 10 '18 at 17:53
Jerry DamsonJerry Damson
111
edited Dec 10 '18 at 18:00
Key Flex
7,84461233
asked Dec 10 '18 at 17:53
Jerry DamsonJerry Damson
111
edited Dec 10 '18 at 18:00
Key Flex
7,84461233
edited Dec 10 '18 at 18:00
Key Flex
7,84461233
edited Dec 10 '18 at 18:00
Key Flex
7,84461233
7,84461233
asked Dec 10 '18 at 17:53
Jerry DamsonJerry Damson
111
asked Dec 10 '18 at 17:53
Jerry DamsonJerry Damson
111
asked Dec 10 '18 at 17:53
Jerry DamsonJerry Damson
111
111
1
$begingroup$
You should add a term $ab$ and make it an absurd expression.
$endgroup$
– T. Bongers
Dec 10 '18 at 17:53
2
$begingroup$
What do you get when you multiply out everything under the radical (surd)? Can you factor it?
$endgroup$
– Doug M
Dec 10 '18 at 17:58
add a comment |
1
$begingroup$
You should add a term $ab$ and make it an absurd expression.
$endgroup$
– T. Bongers
Dec 10 '18 at 17:53
2
$begingroup$
What do you get when you multiply out everything under the radical (surd)? Can you factor it?
$endgroup$
– Doug M
Dec 10 '18 at 17:58
1
1
$begingroup$
You should add a term $ab$ and make it an absurd expression.
$endgroup$
– T. Bongers
Dec 10 '18 at 17:53
$begingroup$
You should add a term $ab$ and make it an absurd expression.
$endgroup$
– T. Bongers
Dec 10 '18 at 17:53
2
2
$begingroup$
What do you get when you multiply out everything under the radical (surd)? Can you factor it?
$endgroup$
– Doug M
Dec 10 '18 at 17:58
$begingroup$
What do you get when you multiply out everything under the radical (surd)? Can you factor it?
$endgroup$
– Doug M
Dec 10 '18 at 17:58
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Things are set up nicely to factor as squares; in particular,
$$(a^2 - b^2)^2 + (2ab)^2 = a^4 + b^4 - 2a^2 b^2 + 4a^2 b^2 = a^4 + 2a^2b^2 + b^4 = (a^2 + b^2)^2.$$
This should simplify things reasonably.
$endgroup$
add a comment |
$begingroup$
Hint: How can $x^2+2xy+y^2$ be factored?
answered Dec 10 '18 at 17:57
JamJam
4,98521431
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
Things are set up nicely to factor as squares; in particular,
$$(a^2 - b^2)^2 + (2ab)^2 = a^4 + b^4 - 2a^2 b^2 + 4a^2 b^2 = a^4 + 2a^2b^2 + b^4 = (a^2 + b^2)^2.$$
This should simplify things reasonably.
$endgroup$
add a comment |
$begingroup$
Things are set up nicely to factor as squares; in particular,
$$(a^2 - b^2)^2 + (2ab)^2 = a^4 + b^4 - 2a^2 b^2 + 4a^2 b^2 = a^4 + 2a^2b^2 + b^4 = (a^2 + b^2)^2.$$
This should simplify things reasonably.
$endgroup$
add a comment |
$begingroup$
Things are set up nicely to factor as squares; in particular,
$$(a^2 - b^2)^2 + (2ab)^2 = a^4 + b^4 - 2a^2 b^2 + 4a^2 b^2 = a^4 + 2a^2b^2 + b^4 = (a^2 + b^2)^2.$$
This should simplify things reasonably.
$endgroup$
Things are set up nicely to factor as squares; in particular,
$$(a^2 - b^2)^2 + (2ab)^2 = a^4 + b^4 - 2a^2 b^2 + 4a^2 b^2 = a^4 + 2a^2b^2 + b^4 = (a^2 + b^2)^2.$$
This should simplify things reasonably.
answered Dec 10 '18 at 17:57
community wiki
T. Bongers
add a comment |
add a comment |
$begingroup$
Hint: How can $x^2+2xy+y^2$ be factored?
answered Dec 10 '18 at 17:57
JamJam
4,98521431
$endgroup$
add a comment |
$begingroup$
Hint: How can $x^2+2xy+y^2$ be factored?
answered Dec 10 '18 at 17:57
JamJam
4,98521431
$endgroup$
add a comment |
$begingroup$
Hint: How can $x^2+2xy+y^2$ be factored?
answered Dec 10 '18 at 17:57
JamJam
4,98521431
$endgroup$
Hint: How can $x^2+2xy+y^2$ be factored?
answered Dec 10 '18 at 17:57
JamJam
4,98521431
answered Dec 10 '18 at 17:57
JamJam
4,98521431
answered Dec 10 '18 at 17:57
JamJam
4,98521431
answered Dec 10 '18 at 17:57
JamJam
4,98521431
4,98521431
add a comment |
add a comment |
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1
$begingroup$
You should add a term $ab$ and make it an absurd expression.
$endgroup$
– T. Bongers
Dec 10 '18 at 17:53
2
$begingroup$
What do you get when you multiply out everything under the radical (surd)? Can you factor it?
$endgroup$
– Doug M
Dec 10 '18 at 17:58