Mixing
Need some help solving this equation
Clash Royale CLAN TAG#URR8PPP
up vote
3
down vote
favorite
The equation is
$$sqrtfrac4-xx+sqrtfracx-4x+1=2-sqrtx^2-12$$
I tried squaring both left side and right side then bringing them to same numerator but got lost from there ... any ideas of how should this be solved?
I got to:
$$2cdotsqrt-frac(x-4)^2x^2+x+4cdotsqrtx^2-12=x^2-8-frac4-xx^2+x$$
radicals radical-equations
edited 34 mins ago
mrtaurho
1,9951621
asked 1 hour ago
alex_c
184
New contributor
alex_c is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
 |Â
show 1 more comment
up vote
3
down vote
favorite
The equation is
$$sqrtfrac4-xx+sqrtfracx-4x+1=2-sqrtx^2-12$$
I tried squaring both left side and right side then bringing them to same numerator but got lost from there ... any ideas of how should this be solved?
I got to:
$$2cdotsqrt-frac(x-4)^2x^2+x+4cdotsqrtx^2-12=x^2-8-frac4-xx^2+x$$
radicals radical-equations
edited 34 mins ago
mrtaurho
1,9951621
asked 1 hour ago
alex_c
184
New contributor
alex_c is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Welcome to MSE. Questions like "Here is the task. Solve it for me!" are poorly received on this site. Therefore try to improve your question with an edit. Improving could consist of providing some context concerning your task or by adding what you have tried so far and where did you struggle :)
– mrtaurho
57 mins ago
Hope it's better now
– alex_c
46 mins ago
Hints: If $x$ is required to be a real value notice the square roots of $(4-x)$ and of $(x-4)$ so that studying the different intervals from $-sqrt12,-1,0,sqrt12,4$ may be of interest...
– Raymond Manzoni
42 mins ago
... and that's why "squaring both sides" is the $boxed??text-th$ thing one must do when he's solving an equation. What number goes in the box?
– Saucy O'Path
41 mins ago
@saucy that is an unfair question because I can't count that high. Most radicalequations require some work before squaring them.
– Oscar Lanzi
38 mins ago
 |Â
show 1 more comment
up vote
3
down vote
favorite
up vote
3
down vote
favorite
The equation is
$$sqrtfrac4-xx+sqrtfracx-4x+1=2-sqrtx^2-12$$
I tried squaring both left side and right side then bringing them to same numerator but got lost from there ... any ideas of how should this be solved?
I got to:
$$2cdotsqrt-frac(x-4)^2x^2+x+4cdotsqrtx^2-12=x^2-8-frac4-xx^2+x$$
radicals radical-equations
edited 34 mins ago
mrtaurho
1,9951621
asked 1 hour ago
alex_c
184
New contributor
alex_c is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
The equation is
$$sqrtfrac4-xx+sqrtfracx-4x+1=2-sqrtx^2-12$$
I tried squaring both left side and right side then bringing them to same numerator but got lost from there ... any ideas of how should this be solved?
I got to:
$$2cdotsqrt-frac(x-4)^2x^2+x+4cdotsqrtx^2-12=x^2-8-frac4-xx^2+x$$
radicals radical-equations
radicals radical-equations
edited 34 mins ago
mrtaurho
1,9951621
asked 1 hour ago
alex_c
184
New contributor
alex_c is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 34 mins ago
mrtaurho
1,9951621
asked 1 hour ago
alex_c
184
New contributor
alex_c is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 34 mins ago
mrtaurho
1,9951621
edited 34 mins ago
mrtaurho
1,9951621
edited 34 mins ago
mrtaurho
1,9951621
1,9951621
asked 1 hour ago
alex_c
184
New contributor
alex_c is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 1 hour ago
alex_c
184
asked 1 hour ago
alex_c
184
184
New contributor
alex_c is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
alex_c is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
alex_c is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Welcome to MSE. Questions like "Here is the task. Solve it for me!" are poorly received on this site. Therefore try to improve your question with an edit. Improving could consist of providing some context concerning your task or by adding what you have tried so far and where did you struggle :)
– mrtaurho
57 mins ago
Hope it's better now
– alex_c
46 mins ago
Hints: If $x$ is required to be a real value notice the square roots of $(4-x)$ and of $(x-4)$ so that studying the different intervals from $-sqrt12,-1,0,sqrt12,4$ may be of interest...
– Raymond Manzoni
42 mins ago
... and that's why "squaring both sides" is the $boxed??text-th$ thing one must do when he's solving an equation. What number goes in the box?
– Saucy O'Path
41 mins ago
@saucy that is an unfair question because I can't count that high. Most radicalequations require some work before squaring them.
– Oscar Lanzi
38 mins ago
 |Â
show 1 more comment
Welcome to MSE. Questions like "Here is the task. Solve it for me!" are poorly received on this site. Therefore try to improve your question with an edit. Improving could consist of providing some context concerning your task or by adding what you have tried so far and where did you struggle :)
– mrtaurho
57 mins ago
Hope it's better now
– alex_c
46 mins ago
Hints: If $x$ is required to be a real value notice the square roots of $(4-x)$ and of $(x-4)$ so that studying the different intervals from $-sqrt12,-1,0,sqrt12,4$ may be of interest...
– Raymond Manzoni
42 mins ago
... and that's why "squaring both sides" is the $boxed??text-th$ thing one must do when he's solving an equation. What number goes in the box?
– Saucy O'Path
41 mins ago
@saucy that is an unfair question because I can't count that high. Most radicalequations require some work before squaring them.
– Oscar Lanzi
38 mins ago
Welcome to MSE. Questions like "Here is the task. Solve it for me!" are poorly received on this site. Therefore try to improve your question with an edit. Improving could consist of providing some context concerning your task or by adding what you have tried so far and where did you struggle :)
– mrtaurho
57 mins ago
Welcome to MSE. Questions like "Here is the task. Solve it for me!" are poorly received on this site. Therefore try to improve your question with an edit. Improving could consist of providing some context concerning your task or by adding what you have tried so far and where did you struggle :)
– mrtaurho
57 mins ago
Hope it's better now
– alex_c
46 mins ago
Hope it's better now
– alex_c
46 mins ago
Hints: If $x$ is required to be a real value notice the square roots of $(4-x)$ and of $(x-4)$ so that studying the different intervals from $-sqrt12,-1,0,sqrt12,4$ may be of interest...
– Raymond Manzoni
42 mins ago
Hints: If $x$ is required to be a real value notice the square roots of $(4-x)$ and of $(x-4)$ so that studying the different intervals from $-sqrt12,-1,0,sqrt12,4$ may be of interest...
– Raymond Manzoni
42 mins ago
... and that's why "squaring both sides" is the $boxed??text-th$ thing one must do when he's solving an equation. What number goes in the box?
– Saucy O'Path
41 mins ago
... and that's why "squaring both sides" is the $boxed??text-th$ thing one must do when he's solving an equation. What number goes in the box?
– Saucy O'Path
41 mins ago
@saucy that is an unfair question because I can't count that high. Most radicalequations require some work before squaring them.
– Oscar Lanzi
38 mins ago
@saucy that is an unfair question because I can't count that high. Most radicalequations require some work before squaring them.
– Oscar Lanzi
38 mins ago
 |Â
show 1 more comment
3 Answers
3
active
oldest
votes
up vote
2
down vote
accepted
Left side exist only iff $$frac4-xxgeq 0iff xin (0,4]$$
and $$fracx-4x+1geq 0iff xin (-infty,-1)cup [4,infty)$$
So the only legitimate value for left side is $4$ which works.
answered 17 mins ago
greedoid
32k114287
add a comment |Â
up vote
1
down vote
Since there are three square roots you will need to square the whole equation at least three times. This is not the best way since it produces false solutions every single time and also the complexity of the equation will increase every time.
Therefore observe the numerator within the square roots of LHS and the square roots on the right
$$sqrtfrac4-xx+sqrtfracx-4x+1=2-sqrtx^2-12$$
Note that for $x=4$ the LHS will be zero hence both numerators will become zero and the denominator will be defined. So we get
$$0=2-sqrt4^2-12Rightarrow 0=2-sqrt16-12Rightarrow 0=0$$
And therefore $x_1=4$ is a solution. Also it is the only real solution (according to WolframAlpha). The other two are given by $x_2/3= -0.889727 pm 0.079797 i$ but honestly speaking I guess you only can get them by squaring over and over again eliminating the false solutions in the end.
answered 41 mins ago
mrtaurho
1,9951621
1
Nah, the exercise is solvable with basic precalculus applied in the correct order. Fact is that the correct order starts by solving inequations.
– Saucy O'Path
38 mins ago
@SaucyO'Path For the case that the OP is only asked to find the real solutions simple observing the equation is all that is needed here. Furthermore I am not sure for myself how to start in the correct way using basis precalculus.
– mrtaurho
33 mins ago
1
@SaucyO'Path what do you think it should be done before squaring?
– alex_c
22 mins ago
The usual things: evaluate the domain of existence and after that, if you decide to square both sides, their signs.
– Saucy O'Path
10 mins ago
add a comment |Â
up vote
1
down vote
Once you identify $x=4$ as a real solution by inspection (mrtaurho's answer), you see that it is the only real solution. For all other $x$, the left side is not real. Check the signs of the radicands on the left side for all other possible cases $x>4, 0<x<4, -1<x<0, x<-1$.
There is one other possibility for a real solution. If one of the radicands on the left is positive and the other negative, you can separately try to fit the two terms on the left side to the real and imaginary parts of the right side. But before investing in heavy computations observe that if $-sqrtx^2-12$ is imaginary at all then (according to the usual principal value definitions) it is a negative number times $i$, and any imaginary radicals on the left would be a positive number times $i$. You can't match the signs by this method.
So the only place to look for a real solution is $x=4$.
answered 15 mins ago
Oscar Lanzi
10.8k11734
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Left side exist only iff $$frac4-xxgeq 0iff xin (0,4]$$
and $$fracx-4x+1geq 0iff xin (-infty,-1)cup [4,infty)$$
So the only legitimate value for left side is $4$ which works.
answered 17 mins ago
greedoid
32k114287
add a comment |Â
up vote
2
down vote
accepted
Left side exist only iff $$frac4-xxgeq 0iff xin (0,4]$$
and $$fracx-4x+1geq 0iff xin (-infty,-1)cup [4,infty)$$
So the only legitimate value for left side is $4$ which works.
answered 17 mins ago
greedoid
32k114287
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Left side exist only iff $$frac4-xxgeq 0iff xin (0,4]$$
and $$fracx-4x+1geq 0iff xin (-infty,-1)cup [4,infty)$$
So the only legitimate value for left side is $4$ which works.
answered 17 mins ago
greedoid
32k114287
Left side exist only iff $$frac4-xxgeq 0iff xin (0,4]$$
and $$fracx-4x+1geq 0iff xin (-infty,-1)cup [4,infty)$$
So the only legitimate value for left side is $4$ which works.
answered 17 mins ago
greedoid
32k114287
edited 10 mins ago
answered 17 mins ago
greedoid
32k114287
answered 17 mins ago
greedoid
32k114287
answered 17 mins ago
greedoid
32k114287
32k114287
add a comment |Â
add a comment |Â
up vote
1
down vote
Since there are three square roots you will need to square the whole equation at least three times. This is not the best way since it produces false solutions every single time and also the complexity of the equation will increase every time.
Therefore observe the numerator within the square roots of LHS and the square roots on the right
$$sqrtfrac4-xx+sqrtfracx-4x+1=2-sqrtx^2-12$$
Note that for $x=4$ the LHS will be zero hence both numerators will become zero and the denominator will be defined. So we get
$$0=2-sqrt4^2-12Rightarrow 0=2-sqrt16-12Rightarrow 0=0$$
And therefore $x_1=4$ is a solution. Also it is the only real solution (according to WolframAlpha). The other two are given by $x_2/3= -0.889727 pm 0.079797 i$ but honestly speaking I guess you only can get them by squaring over and over again eliminating the false solutions in the end.
answered 41 mins ago
mrtaurho
1,9951621
1
Nah, the exercise is solvable with basic precalculus applied in the correct order. Fact is that the correct order starts by solving inequations.
– Saucy O'Path
38 mins ago
@SaucyO'Path For the case that the OP is only asked to find the real solutions simple observing the equation is all that is needed here. Furthermore I am not sure for myself how to start in the correct way using basis precalculus.
– mrtaurho
33 mins ago
1
@SaucyO'Path what do you think it should be done before squaring?
– alex_c
22 mins ago
The usual things: evaluate the domain of existence and after that, if you decide to square both sides, their signs.
– Saucy O'Path
10 mins ago
add a comment |Â
up vote
1
down vote
Since there are three square roots you will need to square the whole equation at least three times. This is not the best way since it produces false solutions every single time and also the complexity of the equation will increase every time.
Therefore observe the numerator within the square roots of LHS and the square roots on the right
$$sqrtfrac4-xx+sqrtfracx-4x+1=2-sqrtx^2-12$$
Note that for $x=4$ the LHS will be zero hence both numerators will become zero and the denominator will be defined. So we get
$$0=2-sqrt4^2-12Rightarrow 0=2-sqrt16-12Rightarrow 0=0$$
And therefore $x_1=4$ is a solution. Also it is the only real solution (according to WolframAlpha). The other two are given by $x_2/3= -0.889727 pm 0.079797 i$ but honestly speaking I guess you only can get them by squaring over and over again eliminating the false solutions in the end.
answered 41 mins ago
mrtaurho
1,9951621
1
Nah, the exercise is solvable with basic precalculus applied in the correct order. Fact is that the correct order starts by solving inequations.
– Saucy O'Path
38 mins ago
@SaucyO'Path For the case that the OP is only asked to find the real solutions simple observing the equation is all that is needed here. Furthermore I am not sure for myself how to start in the correct way using basis precalculus.
– mrtaurho
33 mins ago
1
@SaucyO'Path what do you think it should be done before squaring?
– alex_c
22 mins ago
The usual things: evaluate the domain of existence and after that, if you decide to square both sides, their signs.
– Saucy O'Path
10 mins ago
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Since there are three square roots you will need to square the whole equation at least three times. This is not the best way since it produces false solutions every single time and also the complexity of the equation will increase every time.
Therefore observe the numerator within the square roots of LHS and the square roots on the right
$$sqrtfrac4-xx+sqrtfracx-4x+1=2-sqrtx^2-12$$
Note that for $x=4$ the LHS will be zero hence both numerators will become zero and the denominator will be defined. So we get
$$0=2-sqrt4^2-12Rightarrow 0=2-sqrt16-12Rightarrow 0=0$$
And therefore $x_1=4$ is a solution. Also it is the only real solution (according to WolframAlpha). The other two are given by $x_2/3= -0.889727 pm 0.079797 i$ but honestly speaking I guess you only can get them by squaring over and over again eliminating the false solutions in the end.
answered 41 mins ago
mrtaurho
1,9951621
Since there are three square roots you will need to square the whole equation at least three times. This is not the best way since it produces false solutions every single time and also the complexity of the equation will increase every time.
Therefore observe the numerator within the square roots of LHS and the square roots on the right
$$sqrtfrac4-xx+sqrtfracx-4x+1=2-sqrtx^2-12$$
Note that for $x=4$ the LHS will be zero hence both numerators will become zero and the denominator will be defined. So we get
$$0=2-sqrt4^2-12Rightarrow 0=2-sqrt16-12Rightarrow 0=0$$
And therefore $x_1=4$ is a solution. Also it is the only real solution (according to WolframAlpha). The other two are given by $x_2/3= -0.889727 pm 0.079797 i$ but honestly speaking I guess you only can get them by squaring over and over again eliminating the false solutions in the end.
answered 41 mins ago
mrtaurho
1,9951621
answered 41 mins ago
mrtaurho
1,9951621
answered 41 mins ago
mrtaurho
1,9951621
answered 41 mins ago
mrtaurho
1,9951621
1,9951621
1
Nah, the exercise is solvable with basic precalculus applied in the correct order. Fact is that the correct order starts by solving inequations.
– Saucy O'Path
38 mins ago
@SaucyO'Path For the case that the OP is only asked to find the real solutions simple observing the equation is all that is needed here. Furthermore I am not sure for myself how to start in the correct way using basis precalculus.
– mrtaurho
33 mins ago
1
@SaucyO'Path what do you think it should be done before squaring?
– alex_c
22 mins ago
The usual things: evaluate the domain of existence and after that, if you decide to square both sides, their signs.
– Saucy O'Path
10 mins ago
add a comment |Â
1
Nah, the exercise is solvable with basic precalculus applied in the correct order. Fact is that the correct order starts by solving inequations.
– Saucy O'Path
38 mins ago
@SaucyO'Path For the case that the OP is only asked to find the real solutions simple observing the equation is all that is needed here. Furthermore I am not sure for myself how to start in the correct way using basis precalculus.
– mrtaurho
33 mins ago
1
@SaucyO'Path what do you think it should be done before squaring?
– alex_c
22 mins ago
The usual things: evaluate the domain of existence and after that, if you decide to square both sides, their signs.
– Saucy O'Path
10 mins ago
1
1
Nah, the exercise is solvable with basic precalculus applied in the correct order. Fact is that the correct order starts by solving inequations.
– Saucy O'Path
38 mins ago
Nah, the exercise is solvable with basic precalculus applied in the correct order. Fact is that the correct order starts by solving inequations.
– Saucy O'Path
38 mins ago
@SaucyO'Path For the case that the OP is only asked to find the real solutions simple observing the equation is all that is needed here. Furthermore I am not sure for myself how to start in the correct way using basis precalculus.
– mrtaurho
33 mins ago
@SaucyO'Path For the case that the OP is only asked to find the real solutions simple observing the equation is all that is needed here. Furthermore I am not sure for myself how to start in the correct way using basis precalculus.
– mrtaurho
33 mins ago
1
1
@SaucyO'Path what do you think it should be done before squaring?
– alex_c
22 mins ago
@SaucyO'Path what do you think it should be done before squaring?
– alex_c
22 mins ago
The usual things: evaluate the domain of existence and after that, if you decide to square both sides, their signs.
– Saucy O'Path
10 mins ago
The usual things: evaluate the domain of existence and after that, if you decide to square both sides, their signs.
– Saucy O'Path
10 mins ago
add a comment |Â
up vote
1
down vote
Once you identify $x=4$ as a real solution by inspection (mrtaurho's answer), you see that it is the only real solution. For all other $x$, the left side is not real. Check the signs of the radicands on the left side for all other possible cases $x>4, 0<x<4, -1<x<0, x<-1$.
There is one other possibility for a real solution. If one of the radicands on the left is positive and the other negative, you can separately try to fit the two terms on the left side to the real and imaginary parts of the right side. But before investing in heavy computations observe that if $-sqrtx^2-12$ is imaginary at all then (according to the usual principal value definitions) it is a negative number times $i$, and any imaginary radicals on the left would be a positive number times $i$. You can't match the signs by this method.
So the only place to look for a real solution is $x=4$.
answered 15 mins ago
Oscar Lanzi
10.8k11734
add a comment |Â
up vote
1
down vote
Once you identify $x=4$ as a real solution by inspection (mrtaurho's answer), you see that it is the only real solution. For all other $x$, the left side is not real. Check the signs of the radicands on the left side for all other possible cases $x>4, 0<x<4, -1<x<0, x<-1$.
There is one other possibility for a real solution. If one of the radicands on the left is positive and the other negative, you can separately try to fit the two terms on the left side to the real and imaginary parts of the right side. But before investing in heavy computations observe that if $-sqrtx^2-12$ is imaginary at all then (according to the usual principal value definitions) it is a negative number times $i$, and any imaginary radicals on the left would be a positive number times $i$. You can't match the signs by this method.
So the only place to look for a real solution is $x=4$.
answered 15 mins ago
Oscar Lanzi
10.8k11734
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Once you identify $x=4$ as a real solution by inspection (mrtaurho's answer), you see that it is the only real solution. For all other $x$, the left side is not real. Check the signs of the radicands on the left side for all other possible cases $x>4, 0<x<4, -1<x<0, x<-1$.
There is one other possibility for a real solution. If one of the radicands on the left is positive and the other negative, you can separately try to fit the two terms on the left side to the real and imaginary parts of the right side. But before investing in heavy computations observe that if $-sqrtx^2-12$ is imaginary at all then (according to the usual principal value definitions) it is a negative number times $i$, and any imaginary radicals on the left would be a positive number times $i$. You can't match the signs by this method.
So the only place to look for a real solution is $x=4$.
answered 15 mins ago
Oscar Lanzi
10.8k11734
Once you identify $x=4$ as a real solution by inspection (mrtaurho's answer), you see that it is the only real solution. For all other $x$, the left side is not real. Check the signs of the radicands on the left side for all other possible cases $x>4, 0<x<4, -1<x<0, x<-1$.
There is one other possibility for a real solution. If one of the radicands on the left is positive and the other negative, you can separately try to fit the two terms on the left side to the real and imaginary parts of the right side. But before investing in heavy computations observe that if $-sqrtx^2-12$ is imaginary at all then (according to the usual principal value definitions) it is a negative number times $i$, and any imaginary radicals on the left would be a positive number times $i$. You can't match the signs by this method.
So the only place to look for a real solution is $x=4$.
answered 15 mins ago
Oscar Lanzi
10.8k11734
edited 5 mins ago
answered 15 mins ago
Oscar Lanzi
10.8k11734
answered 15 mins ago
Oscar Lanzi
10.8k11734
answered 15 mins ago
Oscar Lanzi
10.8k11734
10.8k11734
add a comment |Â
add a comment |Â
alex_c is a new contributor. Be nice, and check out our Code of Conduct.
alex_c is a new contributor. Be nice, and check out our Code of Conduct.
alex_c is a new contributor. Be nice, and check out our Code of Conduct.
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Welcome to MSE. Questions like "Here is the task. Solve it for me!" are poorly received on this site. Therefore try to improve your question with an edit. Improving could consist of providing some context concerning your task or by adding what you have tried so far and where did you struggle :)
– mrtaurho
57 mins ago
Hope it's better now
– alex_c
46 mins ago
Hints: If $x$ is required to be a real value notice the square roots of $(4-x)$ and of $(x-4)$ so that studying the different intervals from $-sqrt12,-1,0,sqrt12,4$ may be of interest...
– Raymond Manzoni
42 mins ago
... and that's why "squaring both sides" is the $boxed??text-th$ thing one must do when he's solving an equation. What number goes in the box?
– Saucy O'Path
41 mins ago
@saucy that is an unfair question because I can't count that high. Most radicalequations require some work before squaring them.
– Oscar Lanzi
38 mins ago