Mixing
Problem when convert $sqrt{A}+sqrt{B}=sqrt{C}+sqrt{D}$ to A+B=C+D.
$begingroup$
This is what my lecturer taught me.
If you have $sqrt{A}+sqrt{B}=sqrt{C}+sqrt{D}$
You can easily convert to $A+B=C+D$ or $AB=CD$
And then he gave me an example.
$sqrt{8x+1}+sqrt{3x-5}=sqrt{7x+4}+sqrt{2x-2}$ Find x.
Which is totally work.
After that, I've tried to make my own problem.
But there are some of the problems doesn't work with this technique.
For example, $sqrt{5x+25}+sqrt{x}=sqrt{4x}+sqrt{3x-5}$.
If I use this technique I'll get $x=30$.
But if I solve the problem normally by power 2 both side I'll get $x = frac { 40 } { 11 } + frac { 10 sqrt { 115 } } { 11 }$.
Questions:
Does anyone know how this technique works?
And what're the limitations?
algebra-precalculus examples-counterexamples radicals
edited Jan 30 at 19:36
Michael Rozenberg
109k1896201
asked Jan 30 at 6:49
b.benb.ben
1284
$endgroup$
add a comment |
$begingroup$
This is what my lecturer taught me.
If you have $sqrt{A}+sqrt{B}=sqrt{C}+sqrt{D}$
You can easily convert to $A+B=C+D$ or $AB=CD$
And then he gave me an example.
$sqrt{8x+1}+sqrt{3x-5}=sqrt{7x+4}+sqrt{2x-2}$ Find x.
Which is totally work.
After that, I've tried to make my own problem.
But there are some of the problems doesn't work with this technique.
For example, $sqrt{5x+25}+sqrt{x}=sqrt{4x}+sqrt{3x-5}$.
If I use this technique I'll get $x=30$.
But if I solve the problem normally by power 2 both side I'll get $x = frac { 40 } { 11 } + frac { 10 sqrt { 115 } } { 11 }$.
Questions:
Does anyone know how this technique works?
And what're the limitations?
algebra-precalculus examples-counterexamples radicals
edited Jan 30 at 19:36
Michael Rozenberg
109k1896201
asked Jan 30 at 6:49
b.benb.ben
1284
$endgroup$
2
$begingroup$
This would work if one among $A$ and $B$ is $0$ $textbf{and}$ one among $C$ and $D$ is $0$. OR in general $sqrt{AB}=sqrt{CD}$.
$endgroup$
– Yadati Kiran
Jan 30 at 6:52
$begingroup$
Yes, it will if (A or B is zero) and (C or D is Zero). But he told me you can use with any equation with 4 square root. And It usually works I've done that. But there are some equations doesn't.
$endgroup$
– b.ben
Jan 30 at 6:59
2
$begingroup$
I think it's not very sensible to think that you can just convert the equation with square roots to the equation without square roots, if it most of the time doesn't work. So I don't recommend trying to learn this. Instead, you can think when the equations are equivalent. The main point is that $$ (sqrt{A} + sqrt{B})^2 = |A| + 2sqrt{AB} + |B| neq A +B $$
$endgroup$
– Matti P.
Jan 30 at 7:00
$begingroup$
@MattiP. Totally agreed I shouldn't learn this. I just figured out the example he gave if we recheck the answer we will see that A=C and B=D.
$endgroup$
– b.ben
Jan 30 at 7:25
add a comment |
$begingroup$
This is what my lecturer taught me.
If you have $sqrt{A}+sqrt{B}=sqrt{C}+sqrt{D}$
You can easily convert to $A+B=C+D$ or $AB=CD$
And then he gave me an example.
$sqrt{8x+1}+sqrt{3x-5}=sqrt{7x+4}+sqrt{2x-2}$ Find x.
Which is totally work.
After that, I've tried to make my own problem.
But there are some of the problems doesn't work with this technique.
For example, $sqrt{5x+25}+sqrt{x}=sqrt{4x}+sqrt{3x-5}$.
If I use this technique I'll get $x=30$.
But if I solve the problem normally by power 2 both side I'll get $x = frac { 40 } { 11 } + frac { 10 sqrt { 115 } } { 11 }$.
Questions:
Does anyone know how this technique works?
And what're the limitations?
algebra-precalculus examples-counterexamples radicals
edited Jan 30 at 19:36
Michael Rozenberg
109k1896201
asked Jan 30 at 6:49
b.benb.ben
1284
$endgroup$
This is what my lecturer taught me.
If you have $sqrt{A}+sqrt{B}=sqrt{C}+sqrt{D}$
You can easily convert to $A+B=C+D$ or $AB=CD$
And then he gave me an example.
$sqrt{8x+1}+sqrt{3x-5}=sqrt{7x+4}+sqrt{2x-2}$ Find x.
Which is totally work.
After that, I've tried to make my own problem.
But there are some of the problems doesn't work with this technique.
For example, $sqrt{5x+25}+sqrt{x}=sqrt{4x}+sqrt{3x-5}$.
If I use this technique I'll get $x=30$.
But if I solve the problem normally by power 2 both side I'll get $x = frac { 40 } { 11 } + frac { 10 sqrt { 115 } } { 11 }$.
Questions:
Does anyone know how this technique works?
And what're the limitations?
algebra-precalculus examples-counterexamples radicals
algebra-precalculus examples-counterexamples radicals
edited Jan 30 at 19:36
Michael Rozenberg
109k1896201
asked Jan 30 at 6:49
b.benb.ben
1284
edited Jan 30 at 19:36
Michael Rozenberg
109k1896201
asked Jan 30 at 6:49
b.benb.ben
1284
edited Jan 30 at 19:36
Michael Rozenberg
109k1896201
edited Jan 30 at 19:36
Michael Rozenberg
109k1896201
edited Jan 30 at 19:36
Michael Rozenberg
109k1896201
109k1896201
asked Jan 30 at 6:49
b.benb.ben
1284
asked Jan 30 at 6:49
b.benb.ben
1284
asked Jan 30 at 6:49
b.benb.ben
1284
1284
2
$begingroup$
This would work if one among $A$ and $B$ is $0$ $textbf{and}$ one among $C$ and $D$ is $0$. OR in general $sqrt{AB}=sqrt{CD}$.
$endgroup$
– Yadati Kiran
Jan 30 at 6:52
$begingroup$
Yes, it will if (A or B is zero) and (C or D is Zero). But he told me you can use with any equation with 4 square root. And It usually works I've done that. But there are some equations doesn't.
$endgroup$
– b.ben
Jan 30 at 6:59
2
$begingroup$
I think it's not very sensible to think that you can just convert the equation with square roots to the equation without square roots, if it most of the time doesn't work. So I don't recommend trying to learn this. Instead, you can think when the equations are equivalent. The main point is that $$ (sqrt{A} + sqrt{B})^2 = |A| + 2sqrt{AB} + |B| neq A +B $$
$endgroup$
– Matti P.
Jan 30 at 7:00
$begingroup$
@MattiP. Totally agreed I shouldn't learn this. I just figured out the example he gave if we recheck the answer we will see that A=C and B=D.
$endgroup$
– b.ben
Jan 30 at 7:25
add a comment |
2
$begingroup$
This would work if one among $A$ and $B$ is $0$ $textbf{and}$ one among $C$ and $D$ is $0$. OR in general $sqrt{AB}=sqrt{CD}$.
$endgroup$
– Yadati Kiran
Jan 30 at 6:52
$begingroup$
Yes, it will if (A or B is zero) and (C or D is Zero). But he told me you can use with any equation with 4 square root. And It usually works I've done that. But there are some equations doesn't.
$endgroup$
– b.ben
Jan 30 at 6:59
2
$begingroup$
I think it's not very sensible to think that you can just convert the equation with square roots to the equation without square roots, if it most of the time doesn't work. So I don't recommend trying to learn this. Instead, you can think when the equations are equivalent. The main point is that $$ (sqrt{A} + sqrt{B})^2 = |A| + 2sqrt{AB} + |B| neq A +B $$
$endgroup$
– Matti P.
Jan 30 at 7:00
$begingroup$
@MattiP. Totally agreed I shouldn't learn this. I just figured out the example he gave if we recheck the answer we will see that A=C and B=D.
$endgroup$
– b.ben
Jan 30 at 7:25
2
2
$begingroup$
This would work if one among $A$ and $B$ is $0$ $textbf{and}$ one among $C$ and $D$ is $0$. OR in general $sqrt{AB}=sqrt{CD}$.
$endgroup$
– Yadati Kiran
Jan 30 at 6:52
$begingroup$
This would work if one among $A$ and $B$ is $0$ $textbf{and}$ one among $C$ and $D$ is $0$. OR in general $sqrt{AB}=sqrt{CD}$.
$endgroup$
– Yadati Kiran
Jan 30 at 6:52
$begingroup$
Yes, it will if (A or B is zero) and (C or D is Zero). But he told me you can use with any equation with 4 square root. And It usually works I've done that. But there are some equations doesn't.
$endgroup$
– b.ben
Jan 30 at 6:59
$begingroup$
Yes, it will if (A or B is zero) and (C or D is Zero). But he told me you can use with any equation with 4 square root. And It usually works I've done that. But there are some equations doesn't.
$endgroup$
– b.ben
Jan 30 at 6:59
2
2
$begingroup$
I think it's not very sensible to think that you can just convert the equation with square roots to the equation without square roots, if it most of the time doesn't work. So I don't recommend trying to learn this. Instead, you can think when the equations are equivalent. The main point is that $$ (sqrt{A} + sqrt{B})^2 = |A| + 2sqrt{AB} + |B| neq A +B $$
$endgroup$
– Matti P.
Jan 30 at 7:00
$begingroup$
I think it's not very sensible to think that you can just convert the equation with square roots to the equation without square roots, if it most of the time doesn't work. So I don't recommend trying to learn this. Instead, you can think when the equations are equivalent. The main point is that $$ (sqrt{A} + sqrt{B})^2 = |A| + 2sqrt{AB} + |B| neq A +B $$
$endgroup$
– Matti P.
Jan 30 at 7:00
$begingroup$
@MattiP. Totally agreed I shouldn't learn this. I just figured out the example he gave if we recheck the answer we will see that A=C and B=D.
$endgroup$
– b.ben
Jan 30 at 7:25
$begingroup$
@MattiP. Totally agreed I shouldn't learn this. I just figured out the example he gave if we recheck the answer we will see that A=C and B=D.
$endgroup$
– b.ben
Jan 30 at 7:25
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Try $A=25$, $B=1$ and $C=D=9$.
We see that $$A+Bneq C+D$$ and $$ABneq CD,$$
but $$sqrt{A}+sqrt{B}=sqrt{C}+sqrt{D}.$$
We can try to understand, when it happens.
Firstly, $A$, $B$, $C$ and $D$ are non-negatives.
If $A=B$ and $C=D$ we obtain $A+B=C+D$ and $AB=CD.$
Now, let $A>B$, $C>D$, but $A+B=C+D$.
Thus, after squaring of the both sides we obtain
$$A+B+2sqrt{AB}=C+D+2sqrt{CD},$$ which gives
$$AB=CD.$$
Thus, $$(A+B)^2-4AB=(C+D)^2-4CD$$ or
$$(A-B)^2=(C-D)^2$$ or
$$A-B=C-D,$$ which after summing with $$A+B=C+D$$ gives $$A=C$$ and $$B=D.$$
answered Jan 30 at 6:58
Michael RozenbergMichael Rozenberg
109k1896201
$endgroup$
add a comment |
$begingroup$
Squaring both sides: $sqrt A+sqrt B=sqrt C + sqrt Dimplies A+B+2sqrt{AB}=C+D+2sqrt{CD}$
Therefore the technique works if $A$ or $B=0$ and $C$ or $D=0$.
edited Jan 30 at 7:03
Yadati Kiran
2,1161622
answered Jan 30 at 7:00
abc...abc...
3,237739
$endgroup$
$begingroup$
I've added A=C and B=D, I just figure out that, Every example my lecturer gave. If I recheck the answer it's always A=C and B=D.
$endgroup$
– b.ben
Jan 30 at 7:21
$begingroup$
If $A=0$ or $B=0$ and $C=0$ or $D=0$, then you are left with a term of the form $sqrt{x}=sqrt{y}$. Surely the technique works in this case, but I guess we already knew that. Similary if $A=C$ and $B=D$, we get the form $2sqrt x=2sqrt y$, so there is nothing going on. The interesting case is when $AB=CD$
$endgroup$
– klirk
Jan 30 at 18:30
add a comment |
$begingroup$
If $A+B=C+D$ and $AB=CD$, then $sqrt{AB}=sqrt{CD}$, so that $(sqrt{A}+sqrt{B})^2 = A+2sqrt{AB}+B=C+2sqrt{CD}+D=(sqrt{C}+sqrt{D})^2$.
So if you have an equation of the form $sqrt{A}+sqrt{B}=sqrt{C}+sqrt{D}$, you can form the equations $A+B=C+D$ and $AB=CD$, and try to find a simultaneous solution to both equations, and that will give a solution to the original equation.
It's not enough to find a solution to just one of $A+B=C+D$ and $AB=CD$, as others have pointed out.
answered Jan 30 at 7:22
Michael HartleyMichael Hartley
1,10555
$endgroup$
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
$begingroup$
Try $A=25$, $B=1$ and $C=D=9$.
We see that $$A+Bneq C+D$$ and $$ABneq CD,$$
but $$sqrt{A}+sqrt{B}=sqrt{C}+sqrt{D}.$$
We can try to understand, when it happens.
Firstly, $A$, $B$, $C$ and $D$ are non-negatives.
If $A=B$ and $C=D$ we obtain $A+B=C+D$ and $AB=CD.$
Now, let $A>B$, $C>D$, but $A+B=C+D$.
Thus, after squaring of the both sides we obtain
$$A+B+2sqrt{AB}=C+D+2sqrt{CD},$$ which gives
$$AB=CD.$$
Thus, $$(A+B)^2-4AB=(C+D)^2-4CD$$ or
$$(A-B)^2=(C-D)^2$$ or
$$A-B=C-D,$$ which after summing with $$A+B=C+D$$ gives $$A=C$$ and $$B=D.$$
answered Jan 30 at 6:58
Michael RozenbergMichael Rozenberg
109k1896201
$endgroup$
add a comment |
$begingroup$
Try $A=25$, $B=1$ and $C=D=9$.
We see that $$A+Bneq C+D$$ and $$ABneq CD,$$
but $$sqrt{A}+sqrt{B}=sqrt{C}+sqrt{D}.$$
We can try to understand, when it happens.
Firstly, $A$, $B$, $C$ and $D$ are non-negatives.
If $A=B$ and $C=D$ we obtain $A+B=C+D$ and $AB=CD.$
Now, let $A>B$, $C>D$, but $A+B=C+D$.
Thus, after squaring of the both sides we obtain
$$A+B+2sqrt{AB}=C+D+2sqrt{CD},$$ which gives
$$AB=CD.$$
Thus, $$(A+B)^2-4AB=(C+D)^2-4CD$$ or
$$(A-B)^2=(C-D)^2$$ or
$$A-B=C-D,$$ which after summing with $$A+B=C+D$$ gives $$A=C$$ and $$B=D.$$
answered Jan 30 at 6:58
Michael RozenbergMichael Rozenberg
109k1896201
$endgroup$
add a comment |
$begingroup$
Try $A=25$, $B=1$ and $C=D=9$.
We see that $$A+Bneq C+D$$ and $$ABneq CD,$$
but $$sqrt{A}+sqrt{B}=sqrt{C}+sqrt{D}.$$
We can try to understand, when it happens.
Firstly, $A$, $B$, $C$ and $D$ are non-negatives.
If $A=B$ and $C=D$ we obtain $A+B=C+D$ and $AB=CD.$
Now, let $A>B$, $C>D$, but $A+B=C+D$.
Thus, after squaring of the both sides we obtain
$$A+B+2sqrt{AB}=C+D+2sqrt{CD},$$ which gives
$$AB=CD.$$
Thus, $$(A+B)^2-4AB=(C+D)^2-4CD$$ or
$$(A-B)^2=(C-D)^2$$ or
$$A-B=C-D,$$ which after summing with $$A+B=C+D$$ gives $$A=C$$ and $$B=D.$$
answered Jan 30 at 6:58
Michael RozenbergMichael Rozenberg
109k1896201
$endgroup$
Try $A=25$, $B=1$ and $C=D=9$.
We see that $$A+Bneq C+D$$ and $$ABneq CD,$$
but $$sqrt{A}+sqrt{B}=sqrt{C}+sqrt{D}.$$
We can try to understand, when it happens.
Firstly, $A$, $B$, $C$ and $D$ are non-negatives.
If $A=B$ and $C=D$ we obtain $A+B=C+D$ and $AB=CD.$
Now, let $A>B$, $C>D$, but $A+B=C+D$.
Thus, after squaring of the both sides we obtain
$$A+B+2sqrt{AB}=C+D+2sqrt{CD},$$ which gives
$$AB=CD.$$
Thus, $$(A+B)^2-4AB=(C+D)^2-4CD$$ or
$$(A-B)^2=(C-D)^2$$ or
$$A-B=C-D,$$ which after summing with $$A+B=C+D$$ gives $$A=C$$ and $$B=D.$$
answered Jan 30 at 6:58
Michael RozenbergMichael Rozenberg
109k1896201
edited Jan 30 at 7:33
answered Jan 30 at 6:58
Michael RozenbergMichael Rozenberg
109k1896201
answered Jan 30 at 6:58
Michael RozenbergMichael Rozenberg
109k1896201
answered Jan 30 at 6:58
Michael RozenbergMichael Rozenberg
109k1896201
109k1896201
add a comment |
add a comment |
$begingroup$
Squaring both sides: $sqrt A+sqrt B=sqrt C + sqrt Dimplies A+B+2sqrt{AB}=C+D+2sqrt{CD}$
Therefore the technique works if $A$ or $B=0$ and $C$ or $D=0$.
edited Jan 30 at 7:03
Yadati Kiran
2,1161622
answered Jan 30 at 7:00
abc...abc...
3,237739
$endgroup$
$begingroup$
I've added A=C and B=D, I just figure out that, Every example my lecturer gave. If I recheck the answer it's always A=C and B=D.
$endgroup$
– b.ben
Jan 30 at 7:21
$begingroup$
If $A=0$ or $B=0$ and $C=0$ or $D=0$, then you are left with a term of the form $sqrt{x}=sqrt{y}$. Surely the technique works in this case, but I guess we already knew that. Similary if $A=C$ and $B=D$, we get the form $2sqrt x=2sqrt y$, so there is nothing going on. The interesting case is when $AB=CD$
$endgroup$
– klirk
Jan 30 at 18:30
add a comment |
$begingroup$
Squaring both sides: $sqrt A+sqrt B=sqrt C + sqrt Dimplies A+B+2sqrt{AB}=C+D+2sqrt{CD}$
Therefore the technique works if $A$ or $B=0$ and $C$ or $D=0$.
edited Jan 30 at 7:03
Yadati Kiran
2,1161622
answered Jan 30 at 7:00
abc...abc...
3,237739
$endgroup$
$begingroup$
I've added A=C and B=D, I just figure out that, Every example my lecturer gave. If I recheck the answer it's always A=C and B=D.
$endgroup$
– b.ben
Jan 30 at 7:21
$begingroup$
If $A=0$ or $B=0$ and $C=0$ or $D=0$, then you are left with a term of the form $sqrt{x}=sqrt{y}$. Surely the technique works in this case, but I guess we already knew that. Similary if $A=C$ and $B=D$, we get the form $2sqrt x=2sqrt y$, so there is nothing going on. The interesting case is when $AB=CD$
$endgroup$
– klirk
Jan 30 at 18:30
add a comment |
$begingroup$
Squaring both sides: $sqrt A+sqrt B=sqrt C + sqrt Dimplies A+B+2sqrt{AB}=C+D+2sqrt{CD}$
Therefore the technique works if $A$ or $B=0$ and $C$ or $D=0$.
edited Jan 30 at 7:03
Yadati Kiran
2,1161622
answered Jan 30 at 7:00
abc...abc...
3,237739
$endgroup$
Squaring both sides: $sqrt A+sqrt B=sqrt C + sqrt Dimplies A+B+2sqrt{AB}=C+D+2sqrt{CD}$
Therefore the technique works if $A$ or $B=0$ and $C$ or $D=0$.
edited Jan 30 at 7:03
Yadati Kiran
2,1161622
answered Jan 30 at 7:00
abc...abc...
3,237739
edited Jan 30 at 7:03
Yadati Kiran
2,1161622
edited Jan 30 at 7:03
Yadati Kiran
2,1161622
edited Jan 30 at 7:03
Yadati Kiran
2,1161622
2,1161622
answered Jan 30 at 7:00
abc...abc...
3,237739
answered Jan 30 at 7:00
abc...abc...
3,237739
answered Jan 30 at 7:00
abc...abc...
3,237739
3,237739
$begingroup$
I've added A=C and B=D, I just figure out that, Every example my lecturer gave. If I recheck the answer it's always A=C and B=D.
$endgroup$
– b.ben
Jan 30 at 7:21
$begingroup$
If $A=0$ or $B=0$ and $C=0$ or $D=0$, then you are left with a term of the form $sqrt{x}=sqrt{y}$. Surely the technique works in this case, but I guess we already knew that. Similary if $A=C$ and $B=D$, we get the form $2sqrt x=2sqrt y$, so there is nothing going on. The interesting case is when $AB=CD$
$endgroup$
– klirk
Jan 30 at 18:30
add a comment |
$begingroup$
I've added A=C and B=D, I just figure out that, Every example my lecturer gave. If I recheck the answer it's always A=C and B=D.
$endgroup$
– b.ben
Jan 30 at 7:21
$begingroup$
If $A=0$ or $B=0$ and $C=0$ or $D=0$, then you are left with a term of the form $sqrt{x}=sqrt{y}$. Surely the technique works in this case, but I guess we already knew that. Similary if $A=C$ and $B=D$, we get the form $2sqrt x=2sqrt y$, so there is nothing going on. The interesting case is when $AB=CD$
$endgroup$
– klirk
Jan 30 at 18:30
$begingroup$
I've added A=C and B=D, I just figure out that, Every example my lecturer gave. If I recheck the answer it's always A=C and B=D.
$endgroup$
– b.ben
Jan 30 at 7:21
$begingroup$
I've added A=C and B=D, I just figure out that, Every example my lecturer gave. If I recheck the answer it's always A=C and B=D.
$endgroup$
– b.ben
Jan 30 at 7:21
$begingroup$
If $A=0$ or $B=0$ and $C=0$ or $D=0$, then you are left with a term of the form $sqrt{x}=sqrt{y}$. Surely the technique works in this case, but I guess we already knew that. Similary if $A=C$ and $B=D$, we get the form $2sqrt x=2sqrt y$, so there is nothing going on. The interesting case is when $AB=CD$
$endgroup$
– klirk
Jan 30 at 18:30
$begingroup$
If $A=0$ or $B=0$ and $C=0$ or $D=0$, then you are left with a term of the form $sqrt{x}=sqrt{y}$. Surely the technique works in this case, but I guess we already knew that. Similary if $A=C$ and $B=D$, we get the form $2sqrt x=2sqrt y$, so there is nothing going on. The interesting case is when $AB=CD$
$endgroup$
– klirk
Jan 30 at 18:30
add a comment |
$begingroup$
If $A+B=C+D$ and $AB=CD$, then $sqrt{AB}=sqrt{CD}$, so that $(sqrt{A}+sqrt{B})^2 = A+2sqrt{AB}+B=C+2sqrt{CD}+D=(sqrt{C}+sqrt{D})^2$.
So if you have an equation of the form $sqrt{A}+sqrt{B}=sqrt{C}+sqrt{D}$, you can form the equations $A+B=C+D$ and $AB=CD$, and try to find a simultaneous solution to both equations, and that will give a solution to the original equation.
It's not enough to find a solution to just one of $A+B=C+D$ and $AB=CD$, as others have pointed out.
answered Jan 30 at 7:22
Michael HartleyMichael Hartley
1,10555
$endgroup$
add a comment |
$begingroup$
If $A+B=C+D$ and $AB=CD$, then $sqrt{AB}=sqrt{CD}$, so that $(sqrt{A}+sqrt{B})^2 = A+2sqrt{AB}+B=C+2sqrt{CD}+D=(sqrt{C}+sqrt{D})^2$.
So if you have an equation of the form $sqrt{A}+sqrt{B}=sqrt{C}+sqrt{D}$, you can form the equations $A+B=C+D$ and $AB=CD$, and try to find a simultaneous solution to both equations, and that will give a solution to the original equation.
It's not enough to find a solution to just one of $A+B=C+D$ and $AB=CD$, as others have pointed out.
answered Jan 30 at 7:22
Michael HartleyMichael Hartley
1,10555
$endgroup$
add a comment |
$begingroup$
If $A+B=C+D$ and $AB=CD$, then $sqrt{AB}=sqrt{CD}$, so that $(sqrt{A}+sqrt{B})^2 = A+2sqrt{AB}+B=C+2sqrt{CD}+D=(sqrt{C}+sqrt{D})^2$.
So if you have an equation of the form $sqrt{A}+sqrt{B}=sqrt{C}+sqrt{D}$, you can form the equations $A+B=C+D$ and $AB=CD$, and try to find a simultaneous solution to both equations, and that will give a solution to the original equation.
It's not enough to find a solution to just one of $A+B=C+D$ and $AB=CD$, as others have pointed out.
answered Jan 30 at 7:22
Michael HartleyMichael Hartley
1,10555
$endgroup$
If $A+B=C+D$ and $AB=CD$, then $sqrt{AB}=sqrt{CD}$, so that $(sqrt{A}+sqrt{B})^2 = A+2sqrt{AB}+B=C+2sqrt{CD}+D=(sqrt{C}+sqrt{D})^2$.
So if you have an equation of the form $sqrt{A}+sqrt{B}=sqrt{C}+sqrt{D}$, you can form the equations $A+B=C+D$ and $AB=CD$, and try to find a simultaneous solution to both equations, and that will give a solution to the original equation.
It's not enough to find a solution to just one of $A+B=C+D$ and $AB=CD$, as others have pointed out.
answered Jan 30 at 7:22
Michael HartleyMichael Hartley
1,10555
answered Jan 30 at 7:22
Michael HartleyMichael Hartley
1,10555
answered Jan 30 at 7:22
Michael HartleyMichael Hartley
1,10555
answered Jan 30 at 7:22
Michael HartleyMichael Hartley
1,10555
1,10555
add a comment |
add a comment |
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$begingroup$
This would work if one among $A$ and $B$ is $0$ $textbf{and}$ one among $C$ and $D$ is $0$. OR in general $sqrt{AB}=sqrt{CD}$.
$endgroup$
– Yadati Kiran
Jan 30 at 6:52
$begingroup$
Yes, it will if (A or B is zero) and (C or D is Zero). But he told me you can use with any equation with 4 square root. And It usually works I've done that. But there are some equations doesn't.
$endgroup$
– b.ben
Jan 30 at 6:59
2
$begingroup$
I think it's not very sensible to think that you can just convert the equation with square roots to the equation without square roots, if it most of the time doesn't work. So I don't recommend trying to learn this. Instead, you can think when the equations are equivalent. The main point is that $$ (sqrt{A} + sqrt{B})^2 = |A| + 2sqrt{AB} + |B| neq A +B $$
$endgroup$
– Matti P.
Jan 30 at 7:00
$begingroup$
@MattiP. Totally agreed I shouldn't learn this. I just figured out the example he gave if we recheck the answer we will see that A=C and B=D.
$endgroup$
– b.ben
Jan 30 at 7:25