The quadratic equation $ax^2+bx+c=0$ and $Ax^2+Bx+C$ have roots $(x_1,x_2)$ and $(x_3,x_4)$. Find the condition in which the devide.
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The quadratic equations $ax^2+bx+c=0 $ and $Ax^2+Bx+C=0$ have roots $x_1,x_2$ and $x_3,x_4$. Find the condition that the points $(x_1,0)$ and $(x_2,0)$ divide the segment between $(x_3,0)$ and $(x_4,0)$ harmonically. Given two points, $p$ and $q$ are said to divide $a b$ harmonically if one divides $a b$ internally and the other one divides $a b$ externally in the same ratio, in other words, the sum of the ratios is zero. I tried out this question using Section formula and putting the values but I could not get the desired result please help me out.
algebra-precalculus polynomials quadratics coordinate-systems
edited Aug 25 at 1:06
Cornman
2,70421228
asked Aug 25 at 0:42
user584880
91
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The quadratic equations $ax^2+bx+c=0 $ and $Ax^2+Bx+C=0$ have roots $x_1,x_2$ and $x_3,x_4$. Find the condition that the points $(x_1,0)$ and $(x_2,0)$ divide the segment between $(x_3,0)$ and $(x_4,0)$ harmonically. Given two points, $p$ and $q$ are said to divide $a b$ harmonically if one divides $a b$ internally and the other one divides $a b$ externally in the same ratio, in other words, the sum of the ratios is zero. I tried out this question using Section formula and putting the values but I could not get the desired result please help me out.
algebra-precalculus polynomials quadratics coordinate-systems
edited Aug 25 at 1:06
Cornman
2,70421228
asked Aug 25 at 0:42
user584880
91
Holy cow...either I'm stupid or I couldn't understand 90% of the words here.
– Rushabh Mehta
Aug 25 at 0:44
Under the interpretation that $;dfracx_1-x_3x_1-x_4+dfracx_2-x_3x_2-x_4=0,$, where did you get stuck?
– dxiv
Aug 25 at 0:53
add a comment |Â
up vote
0
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up vote
0
down vote
favorite
The quadratic equations $ax^2+bx+c=0 $ and $Ax^2+Bx+C=0$ have roots $x_1,x_2$ and $x_3,x_4$. Find the condition that the points $(x_1,0)$ and $(x_2,0)$ divide the segment between $(x_3,0)$ and $(x_4,0)$ harmonically. Given two points, $p$ and $q$ are said to divide $a b$ harmonically if one divides $a b$ internally and the other one divides $a b$ externally in the same ratio, in other words, the sum of the ratios is zero. I tried out this question using Section formula and putting the values but I could not get the desired result please help me out.
algebra-precalculus polynomials quadratics coordinate-systems
edited Aug 25 at 1:06
Cornman
2,70421228
asked Aug 25 at 0:42
user584880
91
The quadratic equations $ax^2+bx+c=0 $ and $Ax^2+Bx+C=0$ have roots $x_1,x_2$ and $x_3,x_4$. Find the condition that the points $(x_1,0)$ and $(x_2,0)$ divide the segment between $(x_3,0)$ and $(x_4,0)$ harmonically. Given two points, $p$ and $q$ are said to divide $a b$ harmonically if one divides $a b$ internally and the other one divides $a b$ externally in the same ratio, in other words, the sum of the ratios is zero. I tried out this question using Section formula and putting the values but I could not get the desired result please help me out.
algebra-precalculus polynomials quadratics coordinate-systems
edited Aug 25 at 1:06
Cornman
2,70421228
asked Aug 25 at 0:42
user584880
91
edited Aug 25 at 1:06
Cornman
2,70421228
edited Aug 25 at 1:06
Cornman
2,70421228
edited Aug 25 at 1:06
Cornman
2,70421228
2,70421228
asked Aug 25 at 0:42
user584880
91
asked Aug 25 at 0:42
user584880
91
asked Aug 25 at 0:42
user584880
91
91
Holy cow...either I'm stupid or I couldn't understand 90% of the words here.
– Rushabh Mehta
Aug 25 at 0:44
Under the interpretation that $;dfracx_1-x_3x_1-x_4+dfracx_2-x_3x_2-x_4=0,$, where did you get stuck?
– dxiv
Aug 25 at 0:53
add a comment |Â
Holy cow...either I'm stupid or I couldn't understand 90% of the words here.
– Rushabh Mehta
Aug 25 at 0:44
Under the interpretation that $;dfracx_1-x_3x_1-x_4+dfracx_2-x_3x_2-x_4=0,$, where did you get stuck?
– dxiv
Aug 25 at 0:53
Holy cow...either I'm stupid or I couldn't understand 90% of the words here.
– Rushabh Mehta
Aug 25 at 0:44
Holy cow...either I'm stupid or I couldn't understand 90% of the words here.
– Rushabh Mehta
Aug 25 at 0:44
Under the interpretation that $;dfracx_1-x_3x_1-x_4+dfracx_2-x_3x_2-x_4=0,$, where did you get stuck?
– dxiv
Aug 25 at 0:53
Under the interpretation that $;dfracx_1-x_3x_1-x_4+dfracx_2-x_3x_2-x_4=0,$, where did you get stuck?
– dxiv
Aug 25 at 0:53
add a comment |Â
1 Answer
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We can manipulate the relation given by dxiv as follows:
$fracx_1-x_3x_1-x_4=fracx_3-x_2x_2-x_4$
$x_1x_2 - x_1x_4-x_2x_3 +x_3x_4=x_1x_3-x_1x_2-x_3x_4+x_2x_4$
Rearranging and summing gives:
$2x_1x_2 +2x_3x_4 -x_2(x_3+x_4)-x_1(x_3+x_4)=0$
$x_1x_2=fracca$
$x_3x_4=fracCA$
$x_3+x_4=-fracBA$
$x_1+x_2=-fracba$
Substituting we get:
$2(fracca+fracCA)-fracba.fracBA=0$
Or:
$Atimes c+atimes C =fracb times B2$
answered Aug 25 at 17:55
sirous
931512
What is dxiv in the answer you have written?
– user584880
Aug 26 at 0:43
see second comment. he or she is the user who proposed the relation.
– sirous
Aug 26 at 6:24
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
We can manipulate the relation given by dxiv as follows:
$fracx_1-x_3x_1-x_4=fracx_3-x_2x_2-x_4$
$x_1x_2 - x_1x_4-x_2x_3 +x_3x_4=x_1x_3-x_1x_2-x_3x_4+x_2x_4$
Rearranging and summing gives:
$2x_1x_2 +2x_3x_4 -x_2(x_3+x_4)-x_1(x_3+x_4)=0$
$x_1x_2=fracca$
$x_3x_4=fracCA$
$x_3+x_4=-fracBA$
$x_1+x_2=-fracba$
Substituting we get:
$2(fracca+fracCA)-fracba.fracBA=0$
Or:
$Atimes c+atimes C =fracb times B2$
answered Aug 25 at 17:55
sirous
931512
What is dxiv in the answer you have written?
– user584880
Aug 26 at 0:43
see second comment. he or she is the user who proposed the relation.
– sirous
Aug 26 at 6:24
add a comment |Â
up vote
0
down vote
We can manipulate the relation given by dxiv as follows:
$fracx_1-x_3x_1-x_4=fracx_3-x_2x_2-x_4$
$x_1x_2 - x_1x_4-x_2x_3 +x_3x_4=x_1x_3-x_1x_2-x_3x_4+x_2x_4$
Rearranging and summing gives:
$2x_1x_2 +2x_3x_4 -x_2(x_3+x_4)-x_1(x_3+x_4)=0$
$x_1x_2=fracca$
$x_3x_4=fracCA$
$x_3+x_4=-fracBA$
$x_1+x_2=-fracba$
Substituting we get:
$2(fracca+fracCA)-fracba.fracBA=0$
Or:
$Atimes c+atimes C =fracb times B2$
answered Aug 25 at 17:55
sirous
931512
What is dxiv in the answer you have written?
– user584880
Aug 26 at 0:43
see second comment. he or she is the user who proposed the relation.
– sirous
Aug 26 at 6:24
add a comment |Â
up vote
0
down vote
up vote
0
down vote
We can manipulate the relation given by dxiv as follows:
$fracx_1-x_3x_1-x_4=fracx_3-x_2x_2-x_4$
$x_1x_2 - x_1x_4-x_2x_3 +x_3x_4=x_1x_3-x_1x_2-x_3x_4+x_2x_4$
Rearranging and summing gives:
$2x_1x_2 +2x_3x_4 -x_2(x_3+x_4)-x_1(x_3+x_4)=0$
$x_1x_2=fracca$
$x_3x_4=fracCA$
$x_3+x_4=-fracBA$
$x_1+x_2=-fracba$
Substituting we get:
$2(fracca+fracCA)-fracba.fracBA=0$
Or:
$Atimes c+atimes C =fracb times B2$
answered Aug 25 at 17:55
sirous
931512
We can manipulate the relation given by dxiv as follows:
$fracx_1-x_3x_1-x_4=fracx_3-x_2x_2-x_4$
$x_1x_2 - x_1x_4-x_2x_3 +x_3x_4=x_1x_3-x_1x_2-x_3x_4+x_2x_4$
Rearranging and summing gives:
$2x_1x_2 +2x_3x_4 -x_2(x_3+x_4)-x_1(x_3+x_4)=0$
$x_1x_2=fracca$
$x_3x_4=fracCA$
$x_3+x_4=-fracBA$
$x_1+x_2=-fracba$
Substituting we get:
$2(fracca+fracCA)-fracba.fracBA=0$
Or:
$Atimes c+atimes C =fracb times B2$
answered Aug 25 at 17:55
sirous
931512
answered Aug 25 at 17:55
sirous
931512
answered Aug 25 at 17:55
sirous
931512
answered Aug 25 at 17:55
sirous
931512
931512
What is dxiv in the answer you have written?
– user584880
Aug 26 at 0:43
see second comment. he or she is the user who proposed the relation.
– sirous
Aug 26 at 6:24
add a comment |Â
What is dxiv in the answer you have written?
– user584880
Aug 26 at 0:43
see second comment. he or she is the user who proposed the relation.
– sirous
Aug 26 at 6:24
What is dxiv in the answer you have written?
– user584880
Aug 26 at 0:43
What is dxiv in the answer you have written?
– user584880
Aug 26 at 0:43
see second comment. he or she is the user who proposed the relation.
– sirous
Aug 26 at 6:24
see second comment. he or she is the user who proposed the relation.
– sirous
Aug 26 at 6:24
add a comment |Â
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Holy cow...either I'm stupid or I couldn't understand 90% of the words here.
– Rushabh Mehta
Aug 25 at 0:44
Under the interpretation that $;dfracx_1-x_3x_1-x_4+dfracx_2-x_3x_2-x_4=0,$, where did you get stuck?
– dxiv
Aug 25 at 0:53