Mixing
Elementary Level Algebra Question
Clash Royale CLAN TAG#URR8PPP
up vote
2
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I'm trying to solve the following homework problem:
If $aneq b$, $a^3-b^3 = 19x^3$ and $a-b=x$, which of the following conclusions is correct?
beginalign
text(1) & a = 3x \
text(2) & a = 3x text or a = -2x \
text(3) & a = -3x text or a = 2x \
text(4) & a = 3x text or a = 2x
endalign
I am getting option (4): $a = 3x$ or $a=2x$ as the answer, which is incorrect. The correct answer is $a=3x$ or $a=-2x.$
My work:-
$$ (a-b)^3 + 3ab(a-b)=19x^3$$
$$x^3 + 3ab(x) = 19x^3$$
$$ab=6x^2$$
Hence by comparing we have,
$$a * b = 3x * 2x text or a * b = 2x * 3x$$
So a can be either $2x$ or $3x$.
Why is my answer wrong?
algebra-precalculus
edited 2 hours ago
user1551
69.2k566124
asked 3 hours ago
Navneet Kumar
356316
add a comment |Â
up vote
2
down vote
favorite
I'm trying to solve the following homework problem:
If $aneq b$, $a^3-b^3 = 19x^3$ and $a-b=x$, which of the following conclusions is correct?
beginalign
text(1) & a = 3x \
text(2) & a = 3x text or a = -2x \
text(3) & a = -3x text or a = 2x \
text(4) & a = 3x text or a = 2x
endalign
I am getting option (4): $a = 3x$ or $a=2x$ as the answer, which is incorrect. The correct answer is $a=3x$ or $a=-2x.$
My work:-
$$ (a-b)^3 + 3ab(a-b)=19x^3$$
$$x^3 + 3ab(x) = 19x^3$$
$$ab=6x^2$$
Hence by comparing we have,
$$a * b = 3x * 2x text or a * b = 2x * 3x$$
So a can be either $2x$ or $3x$.
Why is my answer wrong?
algebra-precalculus
edited 2 hours ago
user1551
69.2k566124
asked 3 hours ago
Navneet Kumar
356316
Please clear up your question. Even without using Latex, it is completely unclear what option (4) or option 2 are, what the premise of the problem is, etc. For your equations, all you need for Latex is start a new line and wrap your math with$$math here$$.
– Steve Heim
3 hours ago
@SteveHeim I am referring to the options in the question.
– Navneet Kumar
3 hours ago
1
Ah I see it now. You should transcribe your question to math.SE instead of putting a link, which will eventually be inaccessible. It also makes your question much more readable.
– Steve Heim
3 hours ago
I edited the latex of your question. You should be able to edit your own answer to see how it looks once it has been applied.
– Steve Heim
3 hours ago
1
Why does $ab=6x^2$ split in the way you describe? There are lots of products that give you $6x^2$, for instance $1/2$ times $12x^2$, $-2x$ times $-3x$ (and infinitely more options). This is where your error arises, try, instead to substitute for $b$ at this point.
– Michael Burr
3 hours ago
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I'm trying to solve the following homework problem:
If $aneq b$, $a^3-b^3 = 19x^3$ and $a-b=x$, which of the following conclusions is correct?
beginalign
text(1) & a = 3x \
text(2) & a = 3x text or a = -2x \
text(3) & a = -3x text or a = 2x \
text(4) & a = 3x text or a = 2x
endalign
I am getting option (4): $a = 3x$ or $a=2x$ as the answer, which is incorrect. The correct answer is $a=3x$ or $a=-2x.$
My work:-
$$ (a-b)^3 + 3ab(a-b)=19x^3$$
$$x^3 + 3ab(x) = 19x^3$$
$$ab=6x^2$$
Hence by comparing we have,
$$a * b = 3x * 2x text or a * b = 2x * 3x$$
So a can be either $2x$ or $3x$.
Why is my answer wrong?
algebra-precalculus
edited 2 hours ago
user1551
69.2k566124
asked 3 hours ago
Navneet Kumar
356316
I'm trying to solve the following homework problem:
If $aneq b$, $a^3-b^3 = 19x^3$ and $a-b=x$, which of the following conclusions is correct?
beginalign
text(1) & a = 3x \
text(2) & a = 3x text or a = -2x \
text(3) & a = -3x text or a = 2x \
text(4) & a = 3x text or a = 2x
endalign
I am getting option (4): $a = 3x$ or $a=2x$ as the answer, which is incorrect. The correct answer is $a=3x$ or $a=-2x.$
My work:-
$$ (a-b)^3 + 3ab(a-b)=19x^3$$
$$x^3 + 3ab(x) = 19x^3$$
$$ab=6x^2$$
Hence by comparing we have,
$$a * b = 3x * 2x text or a * b = 2x * 3x$$
So a can be either $2x$ or $3x$.
Why is my answer wrong?
algebra-precalculus
algebra-precalculus
edited 2 hours ago
user1551
69.2k566124
asked 3 hours ago
Navneet Kumar
356316
edited 2 hours ago
user1551
69.2k566124
asked 3 hours ago
Navneet Kumar
356316
edited 2 hours ago
user1551
69.2k566124
edited 2 hours ago
user1551
69.2k566124
edited 2 hours ago
user1551
69.2k566124
69.2k566124
asked 3 hours ago
Navneet Kumar
356316
asked 3 hours ago
Navneet Kumar
356316
asked 3 hours ago
Navneet Kumar
356316
356316
Please clear up your question. Even without using Latex, it is completely unclear what option (4) or option 2 are, what the premise of the problem is, etc. For your equations, all you need for Latex is start a new line and wrap your math with$$math here$$.
– Steve Heim
3 hours ago
@SteveHeim I am referring to the options in the question.
– Navneet Kumar
3 hours ago
1
Ah I see it now. You should transcribe your question to math.SE instead of putting a link, which will eventually be inaccessible. It also makes your question much more readable.
– Steve Heim
3 hours ago
I edited the latex of your question. You should be able to edit your own answer to see how it looks once it has been applied.
– Steve Heim
3 hours ago
1
Why does $ab=6x^2$ split in the way you describe? There are lots of products that give you $6x^2$, for instance $1/2$ times $12x^2$, $-2x$ times $-3x$ (and infinitely more options). This is where your error arises, try, instead to substitute for $b$ at this point.
– Michael Burr
3 hours ago
add a comment |Â
Please clear up your question. Even without using Latex, it is completely unclear what option (4) or option 2 are, what the premise of the problem is, etc. For your equations, all you need for Latex is start a new line and wrap your math with$$math here$$.
– Steve Heim
3 hours ago
@SteveHeim I am referring to the options in the question.
– Navneet Kumar
3 hours ago
1
Ah I see it now. You should transcribe your question to math.SE instead of putting a link, which will eventually be inaccessible. It also makes your question much more readable.
– Steve Heim
3 hours ago
I edited the latex of your question. You should be able to edit your own answer to see how it looks once it has been applied.
– Steve Heim
3 hours ago
1
Why does $ab=6x^2$ split in the way you describe? There are lots of products that give you $6x^2$, for instance $1/2$ times $12x^2$, $-2x$ times $-3x$ (and infinitely more options). This is where your error arises, try, instead to substitute for $b$ at this point.
– Michael Burr
3 hours ago
Please clear up your question. Even without using Latex, it is completely unclear what option (4) or option 2 are, what the premise of the problem is, etc. For your equations, all you need for Latex is start a new line and wrap your math with
$$ math here $$.– Steve Heim
3 hours ago
Please clear up your question. Even without using Latex, it is completely unclear what option (4) or option 2 are, what the premise of the problem is, etc. For your equations, all you need for Latex is start a new line and wrap your math with
$$ math here $$.– Steve Heim
3 hours ago
@SteveHeim I am referring to the options in the question.
– Navneet Kumar
3 hours ago
@SteveHeim I am referring to the options in the question.
– Navneet Kumar
3 hours ago
1
1
Ah I see it now. You should transcribe your question to math.SE instead of putting a link, which will eventually be inaccessible. It also makes your question much more readable.
– Steve Heim
3 hours ago
Ah I see it now. You should transcribe your question to math.SE instead of putting a link, which will eventually be inaccessible. It also makes your question much more readable.
– Steve Heim
3 hours ago
I edited the latex of your question. You should be able to edit your own answer to see how it looks once it has been applied.
– Steve Heim
3 hours ago
I edited the latex of your question. You should be able to edit your own answer to see how it looks once it has been applied.
– Steve Heim
3 hours ago
1
1
Why does $ab=6x^2$ split in the way you describe? There are lots of products that give you $6x^2$, for instance $1/2$ times $12x^2$, $-2x$ times $-3x$ (and infinitely more options). This is where your error arises, try, instead to substitute for $b$ at this point.
– Michael Burr
3 hours ago
Why does $ab=6x^2$ split in the way you describe? There are lots of products that give you $6x^2$, for instance $1/2$ times $12x^2$, $-2x$ times $-3x$ (and infinitely more options). This is where your error arises, try, instead to substitute for $b$ at this point.
– Michael Burr
3 hours ago
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
2
down vote
accepted
You're losing signs in your division. You actually also have the options
$$ ab = (-3x)(-2x)= 6x^2 text or ab = (-2x)(-3x) = 6x^2 $$
To check, you need to plug all of these into your original constraint. This is easy to do since you know that $ b = a -x$. With $a = 2x$ you get $b=-x$, and if you plug this into your second constraint you get
$$ (2x)^3 - (-x)^3 = 8x^3 + x^3 neq 19x^3 $$
I'll leave it to you to test $a=-2x$.
answered 3 hours ago
Steve Heim
531210
Along with this sign case,I also missed few other cases like (3)(2x^2) or (2)(3x^2). Am I correct?
– Navneet Kumar
3 hours ago
Yes, but these are not relevant for your homework question.
– Steve Heim
3 hours ago
add a comment |Â
up vote
4
down vote
Use difference of cubes.
$$a^3-b^3 = (a-b)(a^2+ab+b^2)$$
$$implies (a-b)(a^2+ab+b^2) = 19x^3$$
Set $colorpurplea-b = x$.
$$implies colorpurplex(a^2+ab+b^2) = 19x^3 implies a^2+ab+b^2 = 19x^2$$
Set $colorblueb = a-x$.
$$implies a^2+acolorblue(a-x)+colorblue(a-x)^2 = 19x^2$$
Move $19x^2$ to the LHS, expand, and simplify.
$$implies a^2+a^2-ax+a^2-2ax+x^2-19x^2 = 0$$
$$implies 3a^2-3ax-18x^2 = 0$$
$$implies a^2-ax-6x^2 = 0$$
Factor the trinomial.
$$implies (a-3x)(a+2x) = 0$$
Set either factor equal to $0$.
$$a = 3x text or a = -2x$$
Edit: You’ve asked where your error is. Your comparison part wasn’t correct.
$$ab = 6x^2$$
You can’t just jump to conclusions on what $a$ and $b$ can be. Here, make the substitution $colorblueb = a-x$.
$$acolorblue(a-x) = 6x^2$$
$$a^2-ax = 6x^2 implies a^2-ax-6x^2 = 0$$
This leads to the same answer.
answered 3 hours ago
KM101
1,148111
Thanks!But,where am I doing error?
– Navneet Kumar
3 hours ago
I’ve edited the answer to include the reason your answer isn’t correct. Michael Burr’s comment sums it up nicely: it’s not possible to just “determine†what $a$ and $b$ can or cannot be. You must carry out substitution to get the answer satisfying all given conditions.
– KM101
3 hours ago
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
You're losing signs in your division. You actually also have the options
$$ ab = (-3x)(-2x)= 6x^2 text or ab = (-2x)(-3x) = 6x^2 $$
To check, you need to plug all of these into your original constraint. This is easy to do since you know that $ b = a -x$. With $a = 2x$ you get $b=-x$, and if you plug this into your second constraint you get
$$ (2x)^3 - (-x)^3 = 8x^3 + x^3 neq 19x^3 $$
I'll leave it to you to test $a=-2x$.
answered 3 hours ago
Steve Heim
531210
Along with this sign case,I also missed few other cases like (3)(2x^2) or (2)(3x^2). Am I correct?
– Navneet Kumar
3 hours ago
Yes, but these are not relevant for your homework question.
– Steve Heim
3 hours ago
add a comment |Â
up vote
2
down vote
accepted
You're losing signs in your division. You actually also have the options
$$ ab = (-3x)(-2x)= 6x^2 text or ab = (-2x)(-3x) = 6x^2 $$
To check, you need to plug all of these into your original constraint. This is easy to do since you know that $ b = a -x$. With $a = 2x$ you get $b=-x$, and if you plug this into your second constraint you get
$$ (2x)^3 - (-x)^3 = 8x^3 + x^3 neq 19x^3 $$
I'll leave it to you to test $a=-2x$.
answered 3 hours ago
Steve Heim
531210
Along with this sign case,I also missed few other cases like (3)(2x^2) or (2)(3x^2). Am I correct?
– Navneet Kumar
3 hours ago
Yes, but these are not relevant for your homework question.
– Steve Heim
3 hours ago
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
You're losing signs in your division. You actually also have the options
$$ ab = (-3x)(-2x)= 6x^2 text or ab = (-2x)(-3x) = 6x^2 $$
To check, you need to plug all of these into your original constraint. This is easy to do since you know that $ b = a -x$. With $a = 2x$ you get $b=-x$, and if you plug this into your second constraint you get
$$ (2x)^3 - (-x)^3 = 8x^3 + x^3 neq 19x^3 $$
I'll leave it to you to test $a=-2x$.
answered 3 hours ago
Steve Heim
531210
You're losing signs in your division. You actually also have the options
$$ ab = (-3x)(-2x)= 6x^2 text or ab = (-2x)(-3x) = 6x^2 $$
To check, you need to plug all of these into your original constraint. This is easy to do since you know that $ b = a -x$. With $a = 2x$ you get $b=-x$, and if you plug this into your second constraint you get
$$ (2x)^3 - (-x)^3 = 8x^3 + x^3 neq 19x^3 $$
I'll leave it to you to test $a=-2x$.
answered 3 hours ago
Steve Heim
531210
answered 3 hours ago
Steve Heim
531210
answered 3 hours ago
Steve Heim
531210
answered 3 hours ago
Steve Heim
531210
531210
Along with this sign case,I also missed few other cases like (3)(2x^2) or (2)(3x^2). Am I correct?
– Navneet Kumar
3 hours ago
Yes, but these are not relevant for your homework question.
– Steve Heim
3 hours ago
add a comment |Â
Along with this sign case,I also missed few other cases like (3)(2x^2) or (2)(3x^2). Am I correct?
– Navneet Kumar
3 hours ago
Yes, but these are not relevant for your homework question.
– Steve Heim
3 hours ago
Along with this sign case,I also missed few other cases like (3)(2x^2) or (2)(3x^2). Am I correct?
– Navneet Kumar
3 hours ago
Along with this sign case,I also missed few other cases like (3)(2x^2) or (2)(3x^2). Am I correct?
– Navneet Kumar
3 hours ago
Yes, but these are not relevant for your homework question.
– Steve Heim
3 hours ago
Yes, but these are not relevant for your homework question.
– Steve Heim
3 hours ago
add a comment |Â
up vote
4
down vote
Use difference of cubes.
$$a^3-b^3 = (a-b)(a^2+ab+b^2)$$
$$implies (a-b)(a^2+ab+b^2) = 19x^3$$
Set $colorpurplea-b = x$.
$$implies colorpurplex(a^2+ab+b^2) = 19x^3 implies a^2+ab+b^2 = 19x^2$$
Set $colorblueb = a-x$.
$$implies a^2+acolorblue(a-x)+colorblue(a-x)^2 = 19x^2$$
Move $19x^2$ to the LHS, expand, and simplify.
$$implies a^2+a^2-ax+a^2-2ax+x^2-19x^2 = 0$$
$$implies 3a^2-3ax-18x^2 = 0$$
$$implies a^2-ax-6x^2 = 0$$
Factor the trinomial.
$$implies (a-3x)(a+2x) = 0$$
Set either factor equal to $0$.
$$a = 3x text or a = -2x$$
Edit: You’ve asked where your error is. Your comparison part wasn’t correct.
$$ab = 6x^2$$
You can’t just jump to conclusions on what $a$ and $b$ can be. Here, make the substitution $colorblueb = a-x$.
$$acolorblue(a-x) = 6x^2$$
$$a^2-ax = 6x^2 implies a^2-ax-6x^2 = 0$$
This leads to the same answer.
answered 3 hours ago
KM101
1,148111
Thanks!But,where am I doing error?
– Navneet Kumar
3 hours ago
I’ve edited the answer to include the reason your answer isn’t correct. Michael Burr’s comment sums it up nicely: it’s not possible to just “determine†what $a$ and $b$ can or cannot be. You must carry out substitution to get the answer satisfying all given conditions.
– KM101
3 hours ago
add a comment |Â
up vote
4
down vote
Use difference of cubes.
$$a^3-b^3 = (a-b)(a^2+ab+b^2)$$
$$implies (a-b)(a^2+ab+b^2) = 19x^3$$
Set $colorpurplea-b = x$.
$$implies colorpurplex(a^2+ab+b^2) = 19x^3 implies a^2+ab+b^2 = 19x^2$$
Set $colorblueb = a-x$.
$$implies a^2+acolorblue(a-x)+colorblue(a-x)^2 = 19x^2$$
Move $19x^2$ to the LHS, expand, and simplify.
$$implies a^2+a^2-ax+a^2-2ax+x^2-19x^2 = 0$$
$$implies 3a^2-3ax-18x^2 = 0$$
$$implies a^2-ax-6x^2 = 0$$
Factor the trinomial.
$$implies (a-3x)(a+2x) = 0$$
Set either factor equal to $0$.
$$a = 3x text or a = -2x$$
Edit: You’ve asked where your error is. Your comparison part wasn’t correct.
$$ab = 6x^2$$
You can’t just jump to conclusions on what $a$ and $b$ can be. Here, make the substitution $colorblueb = a-x$.
$$acolorblue(a-x) = 6x^2$$
$$a^2-ax = 6x^2 implies a^2-ax-6x^2 = 0$$
This leads to the same answer.
answered 3 hours ago
KM101
1,148111
Thanks!But,where am I doing error?
– Navneet Kumar
3 hours ago
I’ve edited the answer to include the reason your answer isn’t correct. Michael Burr’s comment sums it up nicely: it’s not possible to just “determine†what $a$ and $b$ can or cannot be. You must carry out substitution to get the answer satisfying all given conditions.
– KM101
3 hours ago
add a comment |Â
up vote
4
down vote
up vote
4
down vote
Use difference of cubes.
$$a^3-b^3 = (a-b)(a^2+ab+b^2)$$
$$implies (a-b)(a^2+ab+b^2) = 19x^3$$
Set $colorpurplea-b = x$.
$$implies colorpurplex(a^2+ab+b^2) = 19x^3 implies a^2+ab+b^2 = 19x^2$$
Set $colorblueb = a-x$.
$$implies a^2+acolorblue(a-x)+colorblue(a-x)^2 = 19x^2$$
Move $19x^2$ to the LHS, expand, and simplify.
$$implies a^2+a^2-ax+a^2-2ax+x^2-19x^2 = 0$$
$$implies 3a^2-3ax-18x^2 = 0$$
$$implies a^2-ax-6x^2 = 0$$
Factor the trinomial.
$$implies (a-3x)(a+2x) = 0$$
Set either factor equal to $0$.
$$a = 3x text or a = -2x$$
Edit: You’ve asked where your error is. Your comparison part wasn’t correct.
$$ab = 6x^2$$
You can’t just jump to conclusions on what $a$ and $b$ can be. Here, make the substitution $colorblueb = a-x$.
$$acolorblue(a-x) = 6x^2$$
$$a^2-ax = 6x^2 implies a^2-ax-6x^2 = 0$$
This leads to the same answer.
answered 3 hours ago
KM101
1,148111
Use difference of cubes.
$$a^3-b^3 = (a-b)(a^2+ab+b^2)$$
$$implies (a-b)(a^2+ab+b^2) = 19x^3$$
Set $colorpurplea-b = x$.
$$implies colorpurplex(a^2+ab+b^2) = 19x^3 implies a^2+ab+b^2 = 19x^2$$
Set $colorblueb = a-x$.
$$implies a^2+acolorblue(a-x)+colorblue(a-x)^2 = 19x^2$$
Move $19x^2$ to the LHS, expand, and simplify.
$$implies a^2+a^2-ax+a^2-2ax+x^2-19x^2 = 0$$
$$implies 3a^2-3ax-18x^2 = 0$$
$$implies a^2-ax-6x^2 = 0$$
Factor the trinomial.
$$implies (a-3x)(a+2x) = 0$$
Set either factor equal to $0$.
$$a = 3x text or a = -2x$$
Edit: You’ve asked where your error is. Your comparison part wasn’t correct.
$$ab = 6x^2$$
You can’t just jump to conclusions on what $a$ and $b$ can be. Here, make the substitution $colorblueb = a-x$.
$$acolorblue(a-x) = 6x^2$$
$$a^2-ax = 6x^2 implies a^2-ax-6x^2 = 0$$
This leads to the same answer.
answered 3 hours ago
KM101
1,148111
edited 3 hours ago
answered 3 hours ago
KM101
1,148111
answered 3 hours ago
KM101
1,148111
answered 3 hours ago
KM101
1,148111
1,148111
Thanks!But,where am I doing error?
– Navneet Kumar
3 hours ago
I’ve edited the answer to include the reason your answer isn’t correct. Michael Burr’s comment sums it up nicely: it’s not possible to just “determine†what $a$ and $b$ can or cannot be. You must carry out substitution to get the answer satisfying all given conditions.
– KM101
3 hours ago
add a comment |Â
Thanks!But,where am I doing error?
– Navneet Kumar
3 hours ago
I’ve edited the answer to include the reason your answer isn’t correct. Michael Burr’s comment sums it up nicely: it’s not possible to just “determine†what $a$ and $b$ can or cannot be. You must carry out substitution to get the answer satisfying all given conditions.
– KM101
3 hours ago
Thanks!But,where am I doing error?
– Navneet Kumar
3 hours ago
Thanks!But,where am I doing error?
– Navneet Kumar
3 hours ago
I’ve edited the answer to include the reason your answer isn’t correct. Michael Burr’s comment sums it up nicely: it’s not possible to just “determine†what $a$ and $b$ can or cannot be. You must carry out substitution to get the answer satisfying all given conditions.
– KM101
3 hours ago
I’ve edited the answer to include the reason your answer isn’t correct. Michael Burr’s comment sums it up nicely: it’s not possible to just “determine†what $a$ and $b$ can or cannot be. You must carry out substitution to get the answer satisfying all given conditions.
– KM101
3 hours ago
add a comment |Â
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Please clear up your question. Even without using Latex, it is completely unclear what option (4) or option 2 are, what the premise of the problem is, etc. For your equations, all you need for Latex is start a new line and wrap your math with
$$math here$$.– Steve Heim
3 hours ago
@SteveHeim I am referring to the options in the question.
– Navneet Kumar
3 hours ago
1
Ah I see it now. You should transcribe your question to math.SE instead of putting a link, which will eventually be inaccessible. It also makes your question much more readable.
– Steve Heim
3 hours ago
I edited the latex of your question. You should be able to edit your own answer to see how it looks once it has been applied.
– Steve Heim
3 hours ago
1
Why does $ab=6x^2$ split in the way you describe? There are lots of products that give you $6x^2$, for instance $1/2$ times $12x^2$, $-2x$ times $-3x$ (and infinitely more options). This is where your error arises, try, instead to substitute for $b$ at this point.
– Michael Burr
3 hours ago