Why is my solution incorrect for solving these quadratic equations?
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4
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$$frac2x -frac5{sqrt{x}}=1 qquad qquad 10) frac3n -frac7{sqrt{n}} -6=0$$
I have these two problems. For the first one I create a dummy variable,
$y = sqrt x$ then $y^2 = x$.
Substituting this in the first equation, I get:
$displaystyle frac{2}{y^{2}} - frac{5}{y} = 1$
Multiplying both sides by $y^{2}$ I get: $2 - 5y = y^{2}$
So I have $y^{2} +5y-2=0$
Solving for y using completing the square, I get:
$displaystyle y = -frac{5}{2} pm frac{sqrt{33}}{2}$
So I should square this answer to get $x$ since $y^2 = x$
Then my answers are $displaystyle y = frac{58}{4} pm frac{10sqrt{33}}{4}$
But this isn't the correct solution.
Also for $#10$ I do the same thing:
Let $y = sqrt n$ then $y^2 = n$
So I have $displaystyle frac{3}{y^2} - frac{7}{y} -6 = 0$
Multiplying everything by $y^{2}$ gives me $3 -7y - 6y^2 = 0$
So I have $6y^{2} +7y - 3 = 0$
Solving for $y$ using the payback method I get: $displaystyle y = -frac{3}{2}, frac{1}{3}$
Then $n = frac{9}{4}, frac{1}{9}$
But plugging these back in, my solution doesn't work.
I've been doing lots of quadratics using dummy variables and sometimes they work and sometimes they don't.
Here are a list of my problems just so you have some reference:
$$1) (x-7)^2 -13(x-7) +36=0 qquad qquad 4) 3(w/6)^2 -8(w/6) +4=0 \
2) (1-3x)^2 -13(1-3x) +36=0 qquad qquad 5) 3(w^2-2)^2 -8(w^2-2) +4=0 \
3) x^4 -13x^2 +36=0 qquad qquad 6) frac{3}{p^2} -frac{8}{p} +4=0$$
What am I doing wrong and how can I do these sorts of problems using dummy variables?
quadratics
add a comment |
up vote
4
down vote
favorite
$$frac2x -frac5{sqrt{x}}=1 qquad qquad 10) frac3n -frac7{sqrt{n}} -6=0$$
I have these two problems. For the first one I create a dummy variable,
$y = sqrt x$ then $y^2 = x$.
Substituting this in the first equation, I get:
$displaystyle frac{2}{y^{2}} - frac{5}{y} = 1$
Multiplying both sides by $y^{2}$ I get: $2 - 5y = y^{2}$
So I have $y^{2} +5y-2=0$
Solving for y using completing the square, I get:
$displaystyle y = -frac{5}{2} pm frac{sqrt{33}}{2}$
So I should square this answer to get $x$ since $y^2 = x$
Then my answers are $displaystyle y = frac{58}{4} pm frac{10sqrt{33}}{4}$
But this isn't the correct solution.
Also for $#10$ I do the same thing:
Let $y = sqrt n$ then $y^2 = n$
So I have $displaystyle frac{3}{y^2} - frac{7}{y} -6 = 0$
Multiplying everything by $y^{2}$ gives me $3 -7y - 6y^2 = 0$
So I have $6y^{2} +7y - 3 = 0$
Solving for $y$ using the payback method I get: $displaystyle y = -frac{3}{2}, frac{1}{3}$
Then $n = frac{9}{4}, frac{1}{9}$
But plugging these back in, my solution doesn't work.
I've been doing lots of quadratics using dummy variables and sometimes they work and sometimes they don't.
Here are a list of my problems just so you have some reference:
$$1) (x-7)^2 -13(x-7) +36=0 qquad qquad 4) 3(w/6)^2 -8(w/6) +4=0 \
2) (1-3x)^2 -13(1-3x) +36=0 qquad qquad 5) 3(w^2-2)^2 -8(w^2-2) +4=0 \
3) x^4 -13x^2 +36=0 qquad qquad 6) frac{3}{p^2} -frac{8}{p} +4=0$$
What am I doing wrong and how can I do these sorts of problems using dummy variables?
quadratics
This question shows plenty of effort and should thus have more upvotes.
– Shaun
1 hour ago
add a comment |
up vote
4
down vote
favorite
up vote
4
down vote
favorite
$$frac2x -frac5{sqrt{x}}=1 qquad qquad 10) frac3n -frac7{sqrt{n}} -6=0$$
I have these two problems. For the first one I create a dummy variable,
$y = sqrt x$ then $y^2 = x$.
Substituting this in the first equation, I get:
$displaystyle frac{2}{y^{2}} - frac{5}{y} = 1$
Multiplying both sides by $y^{2}$ I get: $2 - 5y = y^{2}$
So I have $y^{2} +5y-2=0$
Solving for y using completing the square, I get:
$displaystyle y = -frac{5}{2} pm frac{sqrt{33}}{2}$
So I should square this answer to get $x$ since $y^2 = x$
Then my answers are $displaystyle y = frac{58}{4} pm frac{10sqrt{33}}{4}$
But this isn't the correct solution.
Also for $#10$ I do the same thing:
Let $y = sqrt n$ then $y^2 = n$
So I have $displaystyle frac{3}{y^2} - frac{7}{y} -6 = 0$
Multiplying everything by $y^{2}$ gives me $3 -7y - 6y^2 = 0$
So I have $6y^{2} +7y - 3 = 0$
Solving for $y$ using the payback method I get: $displaystyle y = -frac{3}{2}, frac{1}{3}$
Then $n = frac{9}{4}, frac{1}{9}$
But plugging these back in, my solution doesn't work.
I've been doing lots of quadratics using dummy variables and sometimes they work and sometimes they don't.
Here are a list of my problems just so you have some reference:
$$1) (x-7)^2 -13(x-7) +36=0 qquad qquad 4) 3(w/6)^2 -8(w/6) +4=0 \
2) (1-3x)^2 -13(1-3x) +36=0 qquad qquad 5) 3(w^2-2)^2 -8(w^2-2) +4=0 \
3) x^4 -13x^2 +36=0 qquad qquad 6) frac{3}{p^2} -frac{8}{p} +4=0$$
What am I doing wrong and how can I do these sorts of problems using dummy variables?
quadratics
$$frac2x -frac5{sqrt{x}}=1 qquad qquad 10) frac3n -frac7{sqrt{n}} -6=0$$
I have these two problems. For the first one I create a dummy variable,
$y = sqrt x$ then $y^2 = x$.
Substituting this in the first equation, I get:
$displaystyle frac{2}{y^{2}} - frac{5}{y} = 1$
Multiplying both sides by $y^{2}$ I get: $2 - 5y = y^{2}$
So I have $y^{2} +5y-2=0$
Solving for y using completing the square, I get:
$displaystyle y = -frac{5}{2} pm frac{sqrt{33}}{2}$
So I should square this answer to get $x$ since $y^2 = x$
Then my answers are $displaystyle y = frac{58}{4} pm frac{10sqrt{33}}{4}$
But this isn't the correct solution.
Also for $#10$ I do the same thing:
Let $y = sqrt n$ then $y^2 = n$
So I have $displaystyle frac{3}{y^2} - frac{7}{y} -6 = 0$
Multiplying everything by $y^{2}$ gives me $3 -7y - 6y^2 = 0$
So I have $6y^{2} +7y - 3 = 0$
Solving for $y$ using the payback method I get: $displaystyle y = -frac{3}{2}, frac{1}{3}$
Then $n = frac{9}{4}, frac{1}{9}$
But plugging these back in, my solution doesn't work.
I've been doing lots of quadratics using dummy variables and sometimes they work and sometimes they don't.
Here are a list of my problems just so you have some reference:
$$1) (x-7)^2 -13(x-7) +36=0 qquad qquad 4) 3(w/6)^2 -8(w/6) +4=0 \
2) (1-3x)^2 -13(1-3x) +36=0 qquad qquad 5) 3(w^2-2)^2 -8(w^2-2) +4=0 \
3) x^4 -13x^2 +36=0 qquad qquad 6) frac{3}{p^2} -frac{8}{p} +4=0$$
What am I doing wrong and how can I do these sorts of problems using dummy variables?
quadratics
quadratics
edited 1 hour ago
KM101
2,254415
2,254415
asked 2 hours ago
user130306
36717
36717
This question shows plenty of effort and should thus have more upvotes.
– Shaun
1 hour ago
add a comment |
This question shows plenty of effort and should thus have more upvotes.
– Shaun
1 hour ago
This question shows plenty of effort and should thus have more upvotes.
– Shaun
1 hour ago
This question shows plenty of effort and should thus have more upvotes.
– Shaun
1 hour ago
add a comment |
4 Answers
4
active
oldest
votes
up vote
2
down vote
accepted
Based on comments some of your issues seemed to be in checking answers and others likely an issue in arithmetic (even the greatest sometimes think $1+1=3$ every now and then, but I can’t speak for them). Here’s a tip I have for checking answers to quadratics.
Generally you have something of the form:
$$Ax^2 + Bx + C = 0$$
You only need to know it is zero. So what I did was this. I didn’t multiply everything. I plugged enough in to verify. So say D = x. I might do the following:
$$DAx + Bx + frac CD x = 0$$
Now you just take the sum $DA + B + frac CD$ which can be done mentally with most high school practice problems.
In the case of $(x-7)^2 - 13(x-7) + 36 = 0$ this makes it much easier to verify $11$ and $16$ as answers.
As for general quadratics your style and technique is good. I don’t see any flaws in your thinking aside from being careful about the back substitutions. Definitely watch for dropping negatives and adding them in. Doing your own new problem might introduce needs to carefully think about the Complex Plane if you aren’t careful. However picking the right substitutions is an art and one I cannot really describe with a process or answer. Most people at your level do not substitutions and so your definitely thinking outside the box. This is a good thing. I did not even know if substitutions until Calculus and it made it very hard to grasp that concept (of just creating a new variable on the fly) so kudos to you.
Your equation "$DAx+Bx+frac{C}{D}x = 0$" makes no sense, so I don't know how you got the (correct) "$DA+B+frac{C}{D} = 0$" out of that.
– user21820
27 mins ago
add a comment |
up vote
3
down vote
You have to pay attention to your domain. In the first equation you get a positive and a negative value for $y$, while $sqrt{x}$, which is your substitution, can only be positive.
even if i choose the positive value for x and try to plug it back in, it isn't the correct answer
– user130306
2 hours ago
Not sure how you checked your answers, but one of them is surely correct, namely the one with the minus in the first equation.
– Makina
2 hours ago
actually I just redid my question 1, (which is on the bottom) and there, there is no restriction on the domain since x can be anything, and i still get teh wrong answer. I get $x =11, 16$ and that doesn't work. I'm referencing $(x-7)^{2}-13(x-7)+36=0$ btw
– user130306
2 hours ago
@user130306 x = 11 is definitely a solution by mental math. Checking 16 now.
– The Great Duck
2 hours ago
so you're saying $x = frac{58}{4} - frac{10sqrt 33}{4}$ works?
– user130306
2 hours ago
|
show 8 more comments
up vote
1
down vote
You are doing your algebra correctly, the only problem is when you have a positive and a negative value for a square root you have to ignore the negative one.
I checked the answer $$ x= frac {58-10 sqrt {33}}{4}$$ for your first problem and it does work nicely.
add a comment |
up vote
0
down vote
In your case, substituting $x = y^2$ might introduce extraneous roots. So you need to check that your computed answers are really answers.
Without any substitution, you could write $dfrac2x -dfrac5{sqrt{x}}=1$ as
$$2 left(dfrac{1}{sqrt x}right)^2 - 5left(dfrac{1}{sqrt x}right)-1 = 0$$
So $$dfrac{1}{sqrt x} = dfrac{5 pm sqrt{33}}{4}$$
We can ignore $dfrac{1}{sqrt x} = dfrac{5 - sqrt{33}}{4}$ since the number on the right side is negative.
So we get $$sqrt x = dfrac{4}{5+sqrt{33}}= dfrac{sqrt{33}-5}{2}$$
and $$x = dfrac{29-5sqrt{33}}{2}$$
Problem $(5)$ for example, can be written as
begin{align}
3color{red}{(w^2-2)}^2 -8color{red}{(w^2-2)} + 4 &= 0 \
(3color{red}{(w^2-2)} - 2)(color{red}{(w^2-2)} - 2) &= 0 \
w^2-2= dfrac 23 &text{ or } w^2-2 = 2 \
w^2 = dfrac 83 &text{ or } w^2 = 4 \
w &in left{dfrac 23 sqrt 6, -dfrac 23 sqrt 6, 2, -2 right}
end{align}
You can solve the others similarly.
add a comment |
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Based on comments some of your issues seemed to be in checking answers and others likely an issue in arithmetic (even the greatest sometimes think $1+1=3$ every now and then, but I can’t speak for them). Here’s a tip I have for checking answers to quadratics.
Generally you have something of the form:
$$Ax^2 + Bx + C = 0$$
You only need to know it is zero. So what I did was this. I didn’t multiply everything. I plugged enough in to verify. So say D = x. I might do the following:
$$DAx + Bx + frac CD x = 0$$
Now you just take the sum $DA + B + frac CD$ which can be done mentally with most high school practice problems.
In the case of $(x-7)^2 - 13(x-7) + 36 = 0$ this makes it much easier to verify $11$ and $16$ as answers.
As for general quadratics your style and technique is good. I don’t see any flaws in your thinking aside from being careful about the back substitutions. Definitely watch for dropping negatives and adding them in. Doing your own new problem might introduce needs to carefully think about the Complex Plane if you aren’t careful. However picking the right substitutions is an art and one I cannot really describe with a process or answer. Most people at your level do not substitutions and so your definitely thinking outside the box. This is a good thing. I did not even know if substitutions until Calculus and it made it very hard to grasp that concept (of just creating a new variable on the fly) so kudos to you.
Your equation "$DAx+Bx+frac{C}{D}x = 0$" makes no sense, so I don't know how you got the (correct) "$DA+B+frac{C}{D} = 0$" out of that.
– user21820
27 mins ago
add a comment |
up vote
2
down vote
accepted
Based on comments some of your issues seemed to be in checking answers and others likely an issue in arithmetic (even the greatest sometimes think $1+1=3$ every now and then, but I can’t speak for them). Here’s a tip I have for checking answers to quadratics.
Generally you have something of the form:
$$Ax^2 + Bx + C = 0$$
You only need to know it is zero. So what I did was this. I didn’t multiply everything. I plugged enough in to verify. So say D = x. I might do the following:
$$DAx + Bx + frac CD x = 0$$
Now you just take the sum $DA + B + frac CD$ which can be done mentally with most high school practice problems.
In the case of $(x-7)^2 - 13(x-7) + 36 = 0$ this makes it much easier to verify $11$ and $16$ as answers.
As for general quadratics your style and technique is good. I don’t see any flaws in your thinking aside from being careful about the back substitutions. Definitely watch for dropping negatives and adding them in. Doing your own new problem might introduce needs to carefully think about the Complex Plane if you aren’t careful. However picking the right substitutions is an art and one I cannot really describe with a process or answer. Most people at your level do not substitutions and so your definitely thinking outside the box. This is a good thing. I did not even know if substitutions until Calculus and it made it very hard to grasp that concept (of just creating a new variable on the fly) so kudos to you.
Your equation "$DAx+Bx+frac{C}{D}x = 0$" makes no sense, so I don't know how you got the (correct) "$DA+B+frac{C}{D} = 0$" out of that.
– user21820
27 mins ago
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Based on comments some of your issues seemed to be in checking answers and others likely an issue in arithmetic (even the greatest sometimes think $1+1=3$ every now and then, but I can’t speak for them). Here’s a tip I have for checking answers to quadratics.
Generally you have something of the form:
$$Ax^2 + Bx + C = 0$$
You only need to know it is zero. So what I did was this. I didn’t multiply everything. I plugged enough in to verify. So say D = x. I might do the following:
$$DAx + Bx + frac CD x = 0$$
Now you just take the sum $DA + B + frac CD$ which can be done mentally with most high school practice problems.
In the case of $(x-7)^2 - 13(x-7) + 36 = 0$ this makes it much easier to verify $11$ and $16$ as answers.
As for general quadratics your style and technique is good. I don’t see any flaws in your thinking aside from being careful about the back substitutions. Definitely watch for dropping negatives and adding them in. Doing your own new problem might introduce needs to carefully think about the Complex Plane if you aren’t careful. However picking the right substitutions is an art and one I cannot really describe with a process or answer. Most people at your level do not substitutions and so your definitely thinking outside the box. This is a good thing. I did not even know if substitutions until Calculus and it made it very hard to grasp that concept (of just creating a new variable on the fly) so kudos to you.
Based on comments some of your issues seemed to be in checking answers and others likely an issue in arithmetic (even the greatest sometimes think $1+1=3$ every now and then, but I can’t speak for them). Here’s a tip I have for checking answers to quadratics.
Generally you have something of the form:
$$Ax^2 + Bx + C = 0$$
You only need to know it is zero. So what I did was this. I didn’t multiply everything. I plugged enough in to verify. So say D = x. I might do the following:
$$DAx + Bx + frac CD x = 0$$
Now you just take the sum $DA + B + frac CD$ which can be done mentally with most high school practice problems.
In the case of $(x-7)^2 - 13(x-7) + 36 = 0$ this makes it much easier to verify $11$ and $16$ as answers.
As for general quadratics your style and technique is good. I don’t see any flaws in your thinking aside from being careful about the back substitutions. Definitely watch for dropping negatives and adding them in. Doing your own new problem might introduce needs to carefully think about the Complex Plane if you aren’t careful. However picking the right substitutions is an art and one I cannot really describe with a process or answer. Most people at your level do not substitutions and so your definitely thinking outside the box. This is a good thing. I did not even know if substitutions until Calculus and it made it very hard to grasp that concept (of just creating a new variable on the fly) so kudos to you.
answered 2 hours ago
The Great Duck
9732047
9732047
Your equation "$DAx+Bx+frac{C}{D}x = 0$" makes no sense, so I don't know how you got the (correct) "$DA+B+frac{C}{D} = 0$" out of that.
– user21820
27 mins ago
add a comment |
Your equation "$DAx+Bx+frac{C}{D}x = 0$" makes no sense, so I don't know how you got the (correct) "$DA+B+frac{C}{D} = 0$" out of that.
– user21820
27 mins ago
Your equation "$DAx+Bx+frac{C}{D}x = 0$" makes no sense, so I don't know how you got the (correct) "$DA+B+frac{C}{D} = 0$" out of that.
– user21820
27 mins ago
Your equation "$DAx+Bx+frac{C}{D}x = 0$" makes no sense, so I don't know how you got the (correct) "$DA+B+frac{C}{D} = 0$" out of that.
– user21820
27 mins ago
add a comment |
up vote
3
down vote
You have to pay attention to your domain. In the first equation you get a positive and a negative value for $y$, while $sqrt{x}$, which is your substitution, can only be positive.
even if i choose the positive value for x and try to plug it back in, it isn't the correct answer
– user130306
2 hours ago
Not sure how you checked your answers, but one of them is surely correct, namely the one with the minus in the first equation.
– Makina
2 hours ago
actually I just redid my question 1, (which is on the bottom) and there, there is no restriction on the domain since x can be anything, and i still get teh wrong answer. I get $x =11, 16$ and that doesn't work. I'm referencing $(x-7)^{2}-13(x-7)+36=0$ btw
– user130306
2 hours ago
@user130306 x = 11 is definitely a solution by mental math. Checking 16 now.
– The Great Duck
2 hours ago
so you're saying $x = frac{58}{4} - frac{10sqrt 33}{4}$ works?
– user130306
2 hours ago
|
show 8 more comments
up vote
3
down vote
You have to pay attention to your domain. In the first equation you get a positive and a negative value for $y$, while $sqrt{x}$, which is your substitution, can only be positive.
even if i choose the positive value for x and try to plug it back in, it isn't the correct answer
– user130306
2 hours ago
Not sure how you checked your answers, but one of them is surely correct, namely the one with the minus in the first equation.
– Makina
2 hours ago
actually I just redid my question 1, (which is on the bottom) and there, there is no restriction on the domain since x can be anything, and i still get teh wrong answer. I get $x =11, 16$ and that doesn't work. I'm referencing $(x-7)^{2}-13(x-7)+36=0$ btw
– user130306
2 hours ago
@user130306 x = 11 is definitely a solution by mental math. Checking 16 now.
– The Great Duck
2 hours ago
so you're saying $x = frac{58}{4} - frac{10sqrt 33}{4}$ works?
– user130306
2 hours ago
|
show 8 more comments
up vote
3
down vote
up vote
3
down vote
You have to pay attention to your domain. In the first equation you get a positive and a negative value for $y$, while $sqrt{x}$, which is your substitution, can only be positive.
You have to pay attention to your domain. In the first equation you get a positive and a negative value for $y$, while $sqrt{x}$, which is your substitution, can only be positive.
answered 2 hours ago
Makina
976113
976113
even if i choose the positive value for x and try to plug it back in, it isn't the correct answer
– user130306
2 hours ago
Not sure how you checked your answers, but one of them is surely correct, namely the one with the minus in the first equation.
– Makina
2 hours ago
actually I just redid my question 1, (which is on the bottom) and there, there is no restriction on the domain since x can be anything, and i still get teh wrong answer. I get $x =11, 16$ and that doesn't work. I'm referencing $(x-7)^{2}-13(x-7)+36=0$ btw
– user130306
2 hours ago
@user130306 x = 11 is definitely a solution by mental math. Checking 16 now.
– The Great Duck
2 hours ago
so you're saying $x = frac{58}{4} - frac{10sqrt 33}{4}$ works?
– user130306
2 hours ago
|
show 8 more comments
even if i choose the positive value for x and try to plug it back in, it isn't the correct answer
– user130306
2 hours ago
Not sure how you checked your answers, but one of them is surely correct, namely the one with the minus in the first equation.
– Makina
2 hours ago
actually I just redid my question 1, (which is on the bottom) and there, there is no restriction on the domain since x can be anything, and i still get teh wrong answer. I get $x =11, 16$ and that doesn't work. I'm referencing $(x-7)^{2}-13(x-7)+36=0$ btw
– user130306
2 hours ago
@user130306 x = 11 is definitely a solution by mental math. Checking 16 now.
– The Great Duck
2 hours ago
so you're saying $x = frac{58}{4} - frac{10sqrt 33}{4}$ works?
– user130306
2 hours ago
even if i choose the positive value for x and try to plug it back in, it isn't the correct answer
– user130306
2 hours ago
even if i choose the positive value for x and try to plug it back in, it isn't the correct answer
– user130306
2 hours ago
Not sure how you checked your answers, but one of them is surely correct, namely the one with the minus in the first equation.
– Makina
2 hours ago
Not sure how you checked your answers, but one of them is surely correct, namely the one with the minus in the first equation.
– Makina
2 hours ago
actually I just redid my question 1, (which is on the bottom) and there, there is no restriction on the domain since x can be anything, and i still get teh wrong answer. I get $x =11, 16$ and that doesn't work. I'm referencing $(x-7)^{2}-13(x-7)+36=0$ btw
– user130306
2 hours ago
actually I just redid my question 1, (which is on the bottom) and there, there is no restriction on the domain since x can be anything, and i still get teh wrong answer. I get $x =11, 16$ and that doesn't work. I'm referencing $(x-7)^{2}-13(x-7)+36=0$ btw
– user130306
2 hours ago
@user130306 x = 11 is definitely a solution by mental math. Checking 16 now.
– The Great Duck
2 hours ago
@user130306 x = 11 is definitely a solution by mental math. Checking 16 now.
– The Great Duck
2 hours ago
so you're saying $x = frac{58}{4} - frac{10sqrt 33}{4}$ works?
– user130306
2 hours ago
so you're saying $x = frac{58}{4} - frac{10sqrt 33}{4}$ works?
– user130306
2 hours ago
|
show 8 more comments
up vote
1
down vote
You are doing your algebra correctly, the only problem is when you have a positive and a negative value for a square root you have to ignore the negative one.
I checked the answer $$ x= frac {58-10 sqrt {33}}{4}$$ for your first problem and it does work nicely.
add a comment |
up vote
1
down vote
You are doing your algebra correctly, the only problem is when you have a positive and a negative value for a square root you have to ignore the negative one.
I checked the answer $$ x= frac {58-10 sqrt {33}}{4}$$ for your first problem and it does work nicely.
add a comment |
up vote
1
down vote
up vote
1
down vote
You are doing your algebra correctly, the only problem is when you have a positive and a negative value for a square root you have to ignore the negative one.
I checked the answer $$ x= frac {58-10 sqrt {33}}{4}$$ for your first problem and it does work nicely.
You are doing your algebra correctly, the only problem is when you have a positive and a negative value for a square root you have to ignore the negative one.
I checked the answer $$ x= frac {58-10 sqrt {33}}{4}$$ for your first problem and it does work nicely.
answered 2 hours ago
Mohammad Riazi-Kermani
40.2k41958
40.2k41958
add a comment |
add a comment |
up vote
0
down vote
In your case, substituting $x = y^2$ might introduce extraneous roots. So you need to check that your computed answers are really answers.
Without any substitution, you could write $dfrac2x -dfrac5{sqrt{x}}=1$ as
$$2 left(dfrac{1}{sqrt x}right)^2 - 5left(dfrac{1}{sqrt x}right)-1 = 0$$
So $$dfrac{1}{sqrt x} = dfrac{5 pm sqrt{33}}{4}$$
We can ignore $dfrac{1}{sqrt x} = dfrac{5 - sqrt{33}}{4}$ since the number on the right side is negative.
So we get $$sqrt x = dfrac{4}{5+sqrt{33}}= dfrac{sqrt{33}-5}{2}$$
and $$x = dfrac{29-5sqrt{33}}{2}$$
Problem $(5)$ for example, can be written as
begin{align}
3color{red}{(w^2-2)}^2 -8color{red}{(w^2-2)} + 4 &= 0 \
(3color{red}{(w^2-2)} - 2)(color{red}{(w^2-2)} - 2) &= 0 \
w^2-2= dfrac 23 &text{ or } w^2-2 = 2 \
w^2 = dfrac 83 &text{ or } w^2 = 4 \
w &in left{dfrac 23 sqrt 6, -dfrac 23 sqrt 6, 2, -2 right}
end{align}
You can solve the others similarly.
add a comment |
up vote
0
down vote
In your case, substituting $x = y^2$ might introduce extraneous roots. So you need to check that your computed answers are really answers.
Without any substitution, you could write $dfrac2x -dfrac5{sqrt{x}}=1$ as
$$2 left(dfrac{1}{sqrt x}right)^2 - 5left(dfrac{1}{sqrt x}right)-1 = 0$$
So $$dfrac{1}{sqrt x} = dfrac{5 pm sqrt{33}}{4}$$
We can ignore $dfrac{1}{sqrt x} = dfrac{5 - sqrt{33}}{4}$ since the number on the right side is negative.
So we get $$sqrt x = dfrac{4}{5+sqrt{33}}= dfrac{sqrt{33}-5}{2}$$
and $$x = dfrac{29-5sqrt{33}}{2}$$
Problem $(5)$ for example, can be written as
begin{align}
3color{red}{(w^2-2)}^2 -8color{red}{(w^2-2)} + 4 &= 0 \
(3color{red}{(w^2-2)} - 2)(color{red}{(w^2-2)} - 2) &= 0 \
w^2-2= dfrac 23 &text{ or } w^2-2 = 2 \
w^2 = dfrac 83 &text{ or } w^2 = 4 \
w &in left{dfrac 23 sqrt 6, -dfrac 23 sqrt 6, 2, -2 right}
end{align}
You can solve the others similarly.
add a comment |
up vote
0
down vote
up vote
0
down vote
In your case, substituting $x = y^2$ might introduce extraneous roots. So you need to check that your computed answers are really answers.
Without any substitution, you could write $dfrac2x -dfrac5{sqrt{x}}=1$ as
$$2 left(dfrac{1}{sqrt x}right)^2 - 5left(dfrac{1}{sqrt x}right)-1 = 0$$
So $$dfrac{1}{sqrt x} = dfrac{5 pm sqrt{33}}{4}$$
We can ignore $dfrac{1}{sqrt x} = dfrac{5 - sqrt{33}}{4}$ since the number on the right side is negative.
So we get $$sqrt x = dfrac{4}{5+sqrt{33}}= dfrac{sqrt{33}-5}{2}$$
and $$x = dfrac{29-5sqrt{33}}{2}$$
Problem $(5)$ for example, can be written as
begin{align}
3color{red}{(w^2-2)}^2 -8color{red}{(w^2-2)} + 4 &= 0 \
(3color{red}{(w^2-2)} - 2)(color{red}{(w^2-2)} - 2) &= 0 \
w^2-2= dfrac 23 &text{ or } w^2-2 = 2 \
w^2 = dfrac 83 &text{ or } w^2 = 4 \
w &in left{dfrac 23 sqrt 6, -dfrac 23 sqrt 6, 2, -2 right}
end{align}
You can solve the others similarly.
In your case, substituting $x = y^2$ might introduce extraneous roots. So you need to check that your computed answers are really answers.
Without any substitution, you could write $dfrac2x -dfrac5{sqrt{x}}=1$ as
$$2 left(dfrac{1}{sqrt x}right)^2 - 5left(dfrac{1}{sqrt x}right)-1 = 0$$
So $$dfrac{1}{sqrt x} = dfrac{5 pm sqrt{33}}{4}$$
We can ignore $dfrac{1}{sqrt x} = dfrac{5 - sqrt{33}}{4}$ since the number on the right side is negative.
So we get $$sqrt x = dfrac{4}{5+sqrt{33}}= dfrac{sqrt{33}-5}{2}$$
and $$x = dfrac{29-5sqrt{33}}{2}$$
Problem $(5)$ for example, can be written as
begin{align}
3color{red}{(w^2-2)}^2 -8color{red}{(w^2-2)} + 4 &= 0 \
(3color{red}{(w^2-2)} - 2)(color{red}{(w^2-2)} - 2) &= 0 \
w^2-2= dfrac 23 &text{ or } w^2-2 = 2 \
w^2 = dfrac 83 &text{ or } w^2 = 4 \
w &in left{dfrac 23 sqrt 6, -dfrac 23 sqrt 6, 2, -2 right}
end{align}
You can solve the others similarly.
edited 46 mins ago
answered 1 hour ago
steven gregory
17.5k22257
17.5k22257
add a comment |
add a comment |
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This question shows plenty of effort and should thus have more upvotes.
– Shaun
1 hour ago