Algebraic Manipulation in the Proof of Heron's formula
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A textbook I'm reading gives a proof of Heron's formula, but has lost me in one of its steps. My mathematical foundations are a bit shaky, so I was hoping someone could explain what was done. The jump I don't understand is going from
$$dfrac{1+dfrac{(a^2+b^2-c^2)}{2ab}}{2}$$
to
$$frac{a^2 + 2ab + b^2 - c^2}{4ab}$$
I tried to simplify the first equation and I ended up with
$$frac{1}{2} + frac{a^2 + b^2 - c^2}{4ab}$$
Is there something I've done wrong? What am I missing?
algebra-precalculus
asked May 14 '15 at 0:55
subjectificationsubjectification
878
$endgroup$
add a comment |
$begingroup$
A textbook I'm reading gives a proof of Heron's formula, but has lost me in one of its steps. My mathematical foundations are a bit shaky, so I was hoping someone could explain what was done. The jump I don't understand is going from
$$dfrac{1+dfrac{(a^2+b^2-c^2)}{2ab}}{2}$$
to
$$frac{a^2 + 2ab + b^2 - c^2}{4ab}$$
I tried to simplify the first equation and I ended up with
$$frac{1}{2} + frac{a^2 + b^2 - c^2}{4ab}$$
Is there something I've done wrong? What am I missing?
algebra-precalculus
asked May 14 '15 at 0:55
subjectificationsubjectification
878
$endgroup$
1
$begingroup$
You don't want to split out the $1$ to make $frac12$, you want to merge it with $frac{a^2 + b^2 + c^2}{2ab}$ to make $frac{a^2 + 2ab + b^2 +c^2}{2ab}$. Other than that, I do think theres a typo regarding $pm c^2$ in there, because there's no legal manipulation that will let you change that sign without changing anything else to compensate.
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– Arthur
May 14 '15 at 0:59
$begingroup$
Yes, good catch - it should be $-c^2$ in both cases.
$endgroup$
– subjectification
May 14 '15 at 1:01
add a comment |
$begingroup$
A textbook I'm reading gives a proof of Heron's formula, but has lost me in one of its steps. My mathematical foundations are a bit shaky, so I was hoping someone could explain what was done. The jump I don't understand is going from
$$dfrac{1+dfrac{(a^2+b^2-c^2)}{2ab}}{2}$$
to
$$frac{a^2 + 2ab + b^2 - c^2}{4ab}$$
I tried to simplify the first equation and I ended up with
$$frac{1}{2} + frac{a^2 + b^2 - c^2}{4ab}$$
Is there something I've done wrong? What am I missing?
algebra-precalculus
asked May 14 '15 at 0:55
subjectificationsubjectification
878
$endgroup$
A textbook I'm reading gives a proof of Heron's formula, but has lost me in one of its steps. My mathematical foundations are a bit shaky, so I was hoping someone could explain what was done. The jump I don't understand is going from
$$dfrac{1+dfrac{(a^2+b^2-c^2)}{2ab}}{2}$$
to
$$frac{a^2 + 2ab + b^2 - c^2}{4ab}$$
I tried to simplify the first equation and I ended up with
$$frac{1}{2} + frac{a^2 + b^2 - c^2}{4ab}$$
Is there something I've done wrong? What am I missing?
algebra-precalculus
algebra-precalculus
asked May 14 '15 at 0:55
subjectificationsubjectification
878
asked May 14 '15 at 0:55
subjectificationsubjectification
878
edited May 14 '15 at 1:02
subjectification
asked May 14 '15 at 0:55
subjectificationsubjectification
878
asked May 14 '15 at 0:55
subjectificationsubjectification
878
asked May 14 '15 at 0:55
subjectificationsubjectification
878
878
1
$begingroup$
You don't want to split out the $1$ to make $frac12$, you want to merge it with $frac{a^2 + b^2 + c^2}{2ab}$ to make $frac{a^2 + 2ab + b^2 +c^2}{2ab}$. Other than that, I do think theres a typo regarding $pm c^2$ in there, because there's no legal manipulation that will let you change that sign without changing anything else to compensate.
$endgroup$
– Arthur
May 14 '15 at 0:59
$begingroup$
Yes, good catch - it should be $-c^2$ in both cases.
$endgroup$
– subjectification
May 14 '15 at 1:01
add a comment |
1
$begingroup$
You don't want to split out the $1$ to make $frac12$, you want to merge it with $frac{a^2 + b^2 + c^2}{2ab}$ to make $frac{a^2 + 2ab + b^2 +c^2}{2ab}$. Other than that, I do think theres a typo regarding $pm c^2$ in there, because there's no legal manipulation that will let you change that sign without changing anything else to compensate.
$endgroup$
– Arthur
May 14 '15 at 0:59
$begingroup$
Yes, good catch - it should be $-c^2$ in both cases.
$endgroup$
– subjectification
May 14 '15 at 1:01
1
1
$begingroup$
You don't want to split out the $1$ to make $frac12$, you want to merge it with $frac{a^2 + b^2 + c^2}{2ab}$ to make $frac{a^2 + 2ab + b^2 +c^2}{2ab}$. Other than that, I do think theres a typo regarding $pm c^2$ in there, because there's no legal manipulation that will let you change that sign without changing anything else to compensate.
$endgroup$
– Arthur
May 14 '15 at 0:59
$begingroup$
You don't want to split out the $1$ to make $frac12$, you want to merge it with $frac{a^2 + b^2 + c^2}{2ab}$ to make $frac{a^2 + 2ab + b^2 +c^2}{2ab}$. Other than that, I do think theres a typo regarding $pm c^2$ in there, because there's no legal manipulation that will let you change that sign without changing anything else to compensate.
$endgroup$
– Arthur
May 14 '15 at 0:59
$begingroup$
Yes, good catch - it should be $-c^2$ in both cases.
$endgroup$
– subjectification
May 14 '15 at 1:01
$begingroup$
Yes, good catch - it should be $-c^2$ in both cases.
$endgroup$
– subjectification
May 14 '15 at 1:01
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Just notice: $$frac12 =frac{2ab}{4ab}$$
So both values are correct.
answered May 14 '15 at 1:00
Thomas AndrewsThomas Andrews
131k12148299
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add a comment |
$begingroup$
$$
dfrac{1+dfrac{(a^2+b^2-c^2)}{2ab}}{2}= dfrac{dfrac{2ab+(a^2+b^2-c^2)}{2ab}}{2} =dfrac{2ab+(a^2+b^2-c^2)}{4ab}.
$$
answered May 14 '15 at 1:02
hjhjhj57hjhjhj57
3,12911233
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add a comment |
$begingroup$
Now further do as,
$frac{1}{2}+frac{a^2+b^2-c^2}{4ab}$
Taking LCM,
$frac{2ab}{4ab}+frac{a^2+b^2-c^2}{4ab}$
$frac{2ab+a^2+b^2-c^2}{4ab}$
edited Jan 10 at 13:00
Abcd
3,19331339
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add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Just notice: $$frac12 =frac{2ab}{4ab}$$
So both values are correct.
answered May 14 '15 at 1:00
Thomas AndrewsThomas Andrews
131k12148299
$endgroup$
add a comment |
$begingroup$
Just notice: $$frac12 =frac{2ab}{4ab}$$
So both values are correct.
answered May 14 '15 at 1:00
Thomas AndrewsThomas Andrews
131k12148299
$endgroup$
add a comment |
$begingroup$
Just notice: $$frac12 =frac{2ab}{4ab}$$
So both values are correct.
answered May 14 '15 at 1:00
Thomas AndrewsThomas Andrews
131k12148299
$endgroup$
Just notice: $$frac12 =frac{2ab}{4ab}$$
So both values are correct.
answered May 14 '15 at 1:00
Thomas AndrewsThomas Andrews
131k12148299
answered May 14 '15 at 1:00
Thomas AndrewsThomas Andrews
131k12148299
answered May 14 '15 at 1:00
Thomas AndrewsThomas Andrews
131k12148299
answered May 14 '15 at 1:00
Thomas AndrewsThomas Andrews
131k12148299
131k12148299
add a comment |
add a comment |
$begingroup$
$$
dfrac{1+dfrac{(a^2+b^2-c^2)}{2ab}}{2}= dfrac{dfrac{2ab+(a^2+b^2-c^2)}{2ab}}{2} =dfrac{2ab+(a^2+b^2-c^2)}{4ab}.
$$
answered May 14 '15 at 1:02
hjhjhj57hjhjhj57
3,12911233
$endgroup$
add a comment |
$begingroup$
$$
dfrac{1+dfrac{(a^2+b^2-c^2)}{2ab}}{2}= dfrac{dfrac{2ab+(a^2+b^2-c^2)}{2ab}}{2} =dfrac{2ab+(a^2+b^2-c^2)}{4ab}.
$$
answered May 14 '15 at 1:02
hjhjhj57hjhjhj57
3,12911233
$endgroup$
add a comment |
$begingroup$
$$
dfrac{1+dfrac{(a^2+b^2-c^2)}{2ab}}{2}= dfrac{dfrac{2ab+(a^2+b^2-c^2)}{2ab}}{2} =dfrac{2ab+(a^2+b^2-c^2)}{4ab}.
$$
answered May 14 '15 at 1:02
hjhjhj57hjhjhj57
3,12911233
$endgroup$
$$
dfrac{1+dfrac{(a^2+b^2-c^2)}{2ab}}{2}= dfrac{dfrac{2ab+(a^2+b^2-c^2)}{2ab}}{2} =dfrac{2ab+(a^2+b^2-c^2)}{4ab}.
$$
answered May 14 '15 at 1:02
hjhjhj57hjhjhj57
3,12911233
answered May 14 '15 at 1:02
hjhjhj57hjhjhj57
3,12911233
answered May 14 '15 at 1:02
hjhjhj57hjhjhj57
3,12911233
answered May 14 '15 at 1:02
hjhjhj57hjhjhj57
3,12911233
3,12911233
add a comment |
add a comment |
$begingroup$
Now further do as,
$frac{1}{2}+frac{a^2+b^2-c^2}{4ab}$
Taking LCM,
$frac{2ab}{4ab}+frac{a^2+b^2-c^2}{4ab}$
$frac{2ab+a^2+b^2-c^2}{4ab}$
edited Jan 10 at 13:00
Abcd
3,19331339
$endgroup$
add a comment |
$begingroup$
Now further do as,
$frac{1}{2}+frac{a^2+b^2-c^2}{4ab}$
Taking LCM,
$frac{2ab}{4ab}+frac{a^2+b^2-c^2}{4ab}$
$frac{2ab+a^2+b^2-c^2}{4ab}$
edited Jan 10 at 13:00
Abcd
3,19331339
$endgroup$
add a comment |
$begingroup$
Now further do as,
$frac{1}{2}+frac{a^2+b^2-c^2}{4ab}$
Taking LCM,
$frac{2ab}{4ab}+frac{a^2+b^2-c^2}{4ab}$
$frac{2ab+a^2+b^2-c^2}{4ab}$
edited Jan 10 at 13:00
Abcd
3,19331339
$endgroup$
Now further do as,
$frac{1}{2}+frac{a^2+b^2-c^2}{4ab}$
Taking LCM,
$frac{2ab}{4ab}+frac{a^2+b^2-c^2}{4ab}$
$frac{2ab+a^2+b^2-c^2}{4ab}$
edited Jan 10 at 13:00
Abcd
3,19331339
edited Jan 10 at 13:00
Abcd
3,19331339
edited Jan 10 at 13:00
Abcd
3,19331339
edited Jan 10 at 13:00
Abcd
3,19331339
3,19331339
answered Jan 9 at 7:10
user629353
add a comment |
add a comment |
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1
$begingroup$
You don't want to split out the $1$ to make $frac12$, you want to merge it with $frac{a^2 + b^2 + c^2}{2ab}$ to make $frac{a^2 + 2ab + b^2 +c^2}{2ab}$. Other than that, I do think theres a typo regarding $pm c^2$ in there, because there's no legal manipulation that will let you change that sign without changing anything else to compensate.
$endgroup$
– Arthur
May 14 '15 at 0:59
$begingroup$
Yes, good catch - it should be $-c^2$ in both cases.
$endgroup$
– subjectification
May 14 '15 at 1:01