I have the following equation to solve analytically: $ 0.5q+sqrt q-2=0$
$begingroup$
I sincerely apologize in advance for the simple question I am asking but I really cannot live with this doubt in my mind.
I have the following equation to solve analytically:
$$ 0.5q+sqrt q -2=0$$
Obviously there are different ways to solve it. If I use the substitution $x=sqrt q$, I can get the solutions to a quadratic-form equation and pick only the positive ones. Then, using the initial definition of $x$, I get the unique solution for $q=1.5279$. However, I also saw a solution which goes as follow:
$$sqrt q =2-0.5q to q=(2-0.5q)^2 to 4+0.25q^2-3q=0,$$
which however yields TWO solutions, $q=1.5279$ and $q=10.47$.
However, by plotting both the initial equation and the last one, I obviously notice that we are dealing with different functions.
Can someone please carefully explain why this happens and if one can actually performs the transformation shown in the last line to find the zeros of the initial function?
algebra-precalculus roots
edited Jan 16 at 19:20
jordan_glen
1
asked Jan 16 at 18:43
MatteoMatteo
154
$endgroup$
add a comment |
$begingroup$
I sincerely apologize in advance for the simple question I am asking but I really cannot live with this doubt in my mind.
I have the following equation to solve analytically:
$$ 0.5q+sqrt q -2=0$$
Obviously there are different ways to solve it. If I use the substitution $x=sqrt q$, I can get the solutions to a quadratic-form equation and pick only the positive ones. Then, using the initial definition of $x$, I get the unique solution for $q=1.5279$. However, I also saw a solution which goes as follow:
$$sqrt q =2-0.5q to q=(2-0.5q)^2 to 4+0.25q^2-3q=0,$$
which however yields TWO solutions, $q=1.5279$ and $q=10.47$.
However, by plotting both the initial equation and the last one, I obviously notice that we are dealing with different functions.
Can someone please carefully explain why this happens and if one can actually performs the transformation shown in the last line to find the zeros of the initial function?
algebra-precalculus roots
edited Jan 16 at 19:20
jordan_glen
1
asked Jan 16 at 18:43
MatteoMatteo
154
$endgroup$
$begingroup$
Remember that squaring always introduces extra solutions.
$endgroup$
– Claude Leibovici
Jan 17 at 6:04
add a comment |
$begingroup$
I sincerely apologize in advance for the simple question I am asking but I really cannot live with this doubt in my mind.
I have the following equation to solve analytically:
$$ 0.5q+sqrt q -2=0$$
Obviously there are different ways to solve it. If I use the substitution $x=sqrt q$, I can get the solutions to a quadratic-form equation and pick only the positive ones. Then, using the initial definition of $x$, I get the unique solution for $q=1.5279$. However, I also saw a solution which goes as follow:
$$sqrt q =2-0.5q to q=(2-0.5q)^2 to 4+0.25q^2-3q=0,$$
which however yields TWO solutions, $q=1.5279$ and $q=10.47$.
However, by plotting both the initial equation and the last one, I obviously notice that we are dealing with different functions.
Can someone please carefully explain why this happens and if one can actually performs the transformation shown in the last line to find the zeros of the initial function?
algebra-precalculus roots
edited Jan 16 at 19:20
jordan_glen
1
asked Jan 16 at 18:43
MatteoMatteo
154
$endgroup$
I sincerely apologize in advance for the simple question I am asking but I really cannot live with this doubt in my mind.
I have the following equation to solve analytically:
$$ 0.5q+sqrt q -2=0$$
Obviously there are different ways to solve it. If I use the substitution $x=sqrt q$, I can get the solutions to a quadratic-form equation and pick only the positive ones. Then, using the initial definition of $x$, I get the unique solution for $q=1.5279$. However, I also saw a solution which goes as follow:
$$sqrt q =2-0.5q to q=(2-0.5q)^2 to 4+0.25q^2-3q=0,$$
which however yields TWO solutions, $q=1.5279$ and $q=10.47$.
However, by plotting both the initial equation and the last one, I obviously notice that we are dealing with different functions.
Can someone please carefully explain why this happens and if one can actually performs the transformation shown in the last line to find the zeros of the initial function?
algebra-precalculus roots
algebra-precalculus roots
edited Jan 16 at 19:20
jordan_glen
1
asked Jan 16 at 18:43
MatteoMatteo
154
edited Jan 16 at 19:20
jordan_glen
1
asked Jan 16 at 18:43
MatteoMatteo
154
edited Jan 16 at 19:20
jordan_glen
1
edited Jan 16 at 19:20
jordan_glen
1
edited Jan 16 at 19:20
jordan_glen
1
1
asked Jan 16 at 18:43
MatteoMatteo
154
asked Jan 16 at 18:43
MatteoMatteo
154
asked Jan 16 at 18:43
MatteoMatteo
154
154
$begingroup$
Remember that squaring always introduces extra solutions.
$endgroup$
– Claude Leibovici
Jan 17 at 6:04
add a comment |
$begingroup$
Remember that squaring always introduces extra solutions.
$endgroup$
– Claude Leibovici
Jan 17 at 6:04
$begingroup$
Remember that squaring always introduces extra solutions.
$endgroup$
– Claude Leibovici
Jan 17 at 6:04
$begingroup$
Remember that squaring always introduces extra solutions.
$endgroup$
– Claude Leibovici
Jan 17 at 6:04
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
However, I also saw a solution which goes as follow:
$q^{1/2}=2-0.5q Rightarrow q=(2-0.5q)^2 Rightarrow 4+0.25q^2-3q=0$
which however yields TWO solutions, $q=1.5279$ and $q=10.47$.
Note that first implication: it says that if $q$ is a solution to the first equation, then it is also a solution to the second. It does not say that all solutions to the second equation are solutions to the first. And indeed, if you take the "bad" solution to the second equation, $q = 10.47$, and look at
$$
0.5 q + q^frac12 - 2
$$
you get a number near $6.4$, so it's definitely not zero.
This is analogous to starting with
$$
q = -2
$$
and squaring both sides to get
$$
q^2 = 4
$$
which has two solutions: $q = 2, q = -2$. But only one of these two solutions is a solution to the original equation.
answered Jan 16 at 18:50
John HughesJohn Hughes
63.9k24191
$endgroup$
add a comment |
$begingroup$
We have the equation $$sqrt{q}=2-frac{1}{2}q$$ with the condition $$4geq qgeq 0$$ since we have a square root on the left-hand side.In such cases it is better to check your result.
answered Jan 16 at 18:46
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
75.1k42865
$endgroup$
add a comment |
$begingroup$
This occurs as it causes the equation to be quadratic, and in this case, it would have two solutions as it is a quadratic. You can use the discriminant to find it out or just (try to) solve it
answered Jan 16 at 20:12
QuaternionQuaternion
1
$endgroup$
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
$begingroup$
However, I also saw a solution which goes as follow:
$q^{1/2}=2-0.5q Rightarrow q=(2-0.5q)^2 Rightarrow 4+0.25q^2-3q=0$
which however yields TWO solutions, $q=1.5279$ and $q=10.47$.
Note that first implication: it says that if $q$ is a solution to the first equation, then it is also a solution to the second. It does not say that all solutions to the second equation are solutions to the first. And indeed, if you take the "bad" solution to the second equation, $q = 10.47$, and look at
$$
0.5 q + q^frac12 - 2
$$
you get a number near $6.4$, so it's definitely not zero.
This is analogous to starting with
$$
q = -2
$$
and squaring both sides to get
$$
q^2 = 4
$$
which has two solutions: $q = 2, q = -2$. But only one of these two solutions is a solution to the original equation.
answered Jan 16 at 18:50
John HughesJohn Hughes
63.9k24191
$endgroup$
add a comment |
$begingroup$
However, I also saw a solution which goes as follow:
$q^{1/2}=2-0.5q Rightarrow q=(2-0.5q)^2 Rightarrow 4+0.25q^2-3q=0$
which however yields TWO solutions, $q=1.5279$ and $q=10.47$.
Note that first implication: it says that if $q$ is a solution to the first equation, then it is also a solution to the second. It does not say that all solutions to the second equation are solutions to the first. And indeed, if you take the "bad" solution to the second equation, $q = 10.47$, and look at
$$
0.5 q + q^frac12 - 2
$$
you get a number near $6.4$, so it's definitely not zero.
This is analogous to starting with
$$
q = -2
$$
and squaring both sides to get
$$
q^2 = 4
$$
which has two solutions: $q = 2, q = -2$. But only one of these two solutions is a solution to the original equation.
answered Jan 16 at 18:50
John HughesJohn Hughes
63.9k24191
$endgroup$
add a comment |
$begingroup$
However, I also saw a solution which goes as follow:
$q^{1/2}=2-0.5q Rightarrow q=(2-0.5q)^2 Rightarrow 4+0.25q^2-3q=0$
which however yields TWO solutions, $q=1.5279$ and $q=10.47$.
Note that first implication: it says that if $q$ is a solution to the first equation, then it is also a solution to the second. It does not say that all solutions to the second equation are solutions to the first. And indeed, if you take the "bad" solution to the second equation, $q = 10.47$, and look at
$$
0.5 q + q^frac12 - 2
$$
you get a number near $6.4$, so it's definitely not zero.
This is analogous to starting with
$$
q = -2
$$
and squaring both sides to get
$$
q^2 = 4
$$
which has two solutions: $q = 2, q = -2$. But only one of these two solutions is a solution to the original equation.
answered Jan 16 at 18:50
John HughesJohn Hughes
63.9k24191
$endgroup$
However, I also saw a solution which goes as follow:
$q^{1/2}=2-0.5q Rightarrow q=(2-0.5q)^2 Rightarrow 4+0.25q^2-3q=0$
which however yields TWO solutions, $q=1.5279$ and $q=10.47$.
Note that first implication: it says that if $q$ is a solution to the first equation, then it is also a solution to the second. It does not say that all solutions to the second equation are solutions to the first. And indeed, if you take the "bad" solution to the second equation, $q = 10.47$, and look at
$$
0.5 q + q^frac12 - 2
$$
you get a number near $6.4$, so it's definitely not zero.
This is analogous to starting with
$$
q = -2
$$
and squaring both sides to get
$$
q^2 = 4
$$
which has two solutions: $q = 2, q = -2$. But only one of these two solutions is a solution to the original equation.
answered Jan 16 at 18:50
John HughesJohn Hughes
63.9k24191
answered Jan 16 at 18:50
John HughesJohn Hughes
63.9k24191
answered Jan 16 at 18:50
John HughesJohn Hughes
63.9k24191
answered Jan 16 at 18:50
John HughesJohn Hughes
63.9k24191
63.9k24191
add a comment |
add a comment |
$begingroup$
We have the equation $$sqrt{q}=2-frac{1}{2}q$$ with the condition $$4geq qgeq 0$$ since we have a square root on the left-hand side.In such cases it is better to check your result.
answered Jan 16 at 18:46
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
75.1k42865
$endgroup$
add a comment |
$begingroup$
We have the equation $$sqrt{q}=2-frac{1}{2}q$$ with the condition $$4geq qgeq 0$$ since we have a square root on the left-hand side.In such cases it is better to check your result.
answered Jan 16 at 18:46
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
75.1k42865
$endgroup$
add a comment |
$begingroup$
We have the equation $$sqrt{q}=2-frac{1}{2}q$$ with the condition $$4geq qgeq 0$$ since we have a square root on the left-hand side.In such cases it is better to check your result.
answered Jan 16 at 18:46
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
75.1k42865
$endgroup$
We have the equation $$sqrt{q}=2-frac{1}{2}q$$ with the condition $$4geq qgeq 0$$ since we have a square root on the left-hand side.In such cases it is better to check your result.
answered Jan 16 at 18:46
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
75.1k42865
answered Jan 16 at 18:46
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
75.1k42865
answered Jan 16 at 18:46
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
75.1k42865
answered Jan 16 at 18:46
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
75.1k42865
75.1k42865
add a comment |
add a comment |
$begingroup$
This occurs as it causes the equation to be quadratic, and in this case, it would have two solutions as it is a quadratic. You can use the discriminant to find it out or just (try to) solve it
answered Jan 16 at 20:12
QuaternionQuaternion
1
$endgroup$
add a comment |
$begingroup$
This occurs as it causes the equation to be quadratic, and in this case, it would have two solutions as it is a quadratic. You can use the discriminant to find it out or just (try to) solve it
answered Jan 16 at 20:12
QuaternionQuaternion
1
$endgroup$
add a comment |
$begingroup$
This occurs as it causes the equation to be quadratic, and in this case, it would have two solutions as it is a quadratic. You can use the discriminant to find it out or just (try to) solve it
answered Jan 16 at 20:12
QuaternionQuaternion
1
$endgroup$
This occurs as it causes the equation to be quadratic, and in this case, it would have two solutions as it is a quadratic. You can use the discriminant to find it out or just (try to) solve it
answered Jan 16 at 20:12
QuaternionQuaternion
1
answered Jan 16 at 20:12
QuaternionQuaternion
1
answered Jan 16 at 20:12
QuaternionQuaternion
1
answered Jan 16 at 20:12
QuaternionQuaternion
1
1
add a comment |
add a comment |
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$begingroup$
Remember that squaring always introduces extra solutions.
$endgroup$
– Claude Leibovici
Jan 17 at 6:04