Square roots in quadratic equation [closed]
$${sqrt x}+ frac{8}{sqrt x} = 6$$
I'm trying to get this to the standard quadratic form so I can factorise it, but not sure how to eliminate the square roots/work around them.
Any help would be appreciated :)
algebra-precalculus quadratics
edited Dec 22 at 12:50
OmG
2,222620
asked Dec 22 at 12:11
Ma Ha
35
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closed as off-topic by TheSimpliFire, Paul Frost, user21820, José Carlos Santos, RRL Dec 22 at 15:53
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – TheSimpliFire, Paul Frost, user21820, José Carlos Santos, RRL
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$${sqrt x}+ frac{8}{sqrt x} = 6$$
I'm trying to get this to the standard quadratic form so I can factorise it, but not sure how to eliminate the square roots/work around them.
Any help would be appreciated :)
algebra-precalculus quadratics
edited Dec 22 at 12:50
OmG
2,222620
asked Dec 22 at 12:11
Ma Ha
35
New contributor
Ma Ha is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
closed as off-topic by TheSimpliFire, Paul Frost, user21820, José Carlos Santos, RRL Dec 22 at 15:53
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – TheSimpliFire, Paul Frost, user21820, José Carlos Santos, RRL
If this question can be reworded to fit the rules in the help center, please edit the question.
1
Substitute $t=sqrt{x}$
– Matko
Dec 22 at 12:12
1
What do you get by squaring both sides?
– Dietrich Burde
Dec 22 at 12:12
Just pay attention to the fact that when you substitute $;t:=sqrt x;$ , the solutions you take must be non-negative, as we have by agreement that $;sqrt xge0;$ .
– DonAntonio
Dec 22 at 12:27
But if it was a cubic, would it be ok to have a negative answer?
– Ma Ha
Dec 22 at 12:31
add a comment |
$${sqrt x}+ frac{8}{sqrt x} = 6$$
I'm trying to get this to the standard quadratic form so I can factorise it, but not sure how to eliminate the square roots/work around them.
Any help would be appreciated :)
algebra-precalculus quadratics
edited Dec 22 at 12:50
OmG
2,222620
asked Dec 22 at 12:11
Ma Ha
35
New contributor
Ma Ha is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$${sqrt x}+ frac{8}{sqrt x} = 6$$
I'm trying to get this to the standard quadratic form so I can factorise it, but not sure how to eliminate the square roots/work around them.
Any help would be appreciated :)
algebra-precalculus quadratics
algebra-precalculus quadratics
edited Dec 22 at 12:50
OmG
2,222620
asked Dec 22 at 12:11
Ma Ha
35
New contributor
Ma Ha is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Dec 22 at 12:50
OmG
2,222620
asked Dec 22 at 12:11
Ma Ha
35
New contributor
Ma Ha is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Dec 22 at 12:50
OmG
2,222620
edited Dec 22 at 12:50
OmG
2,222620
edited Dec 22 at 12:50
OmG
2,222620
2,222620
asked Dec 22 at 12:11
Ma Ha
35
New contributor
Ma Ha is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Dec 22 at 12:11
Ma Ha
35
asked Dec 22 at 12:11
Ma Ha
35
35
New contributor
Ma Ha is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Ma Ha is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Ma Ha is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
closed as off-topic by TheSimpliFire, Paul Frost, user21820, José Carlos Santos, RRL Dec 22 at 15:53
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – TheSimpliFire, Paul Frost, user21820, José Carlos Santos, RRL
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by TheSimpliFire, Paul Frost, user21820, José Carlos Santos, RRL Dec 22 at 15:53
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – TheSimpliFire, Paul Frost, user21820, José Carlos Santos, RRL
If this question can be reworded to fit the rules in the help center, please edit the question.
1
Substitute $t=sqrt{x}$
– Matko
Dec 22 at 12:12
1
What do you get by squaring both sides?
– Dietrich Burde
Dec 22 at 12:12
Just pay attention to the fact that when you substitute $;t:=sqrt x;$ , the solutions you take must be non-negative, as we have by agreement that $;sqrt xge0;$ .
– DonAntonio
Dec 22 at 12:27
But if it was a cubic, would it be ok to have a negative answer?
– Ma Ha
Dec 22 at 12:31
add a comment |
1
Substitute $t=sqrt{x}$
– Matko
Dec 22 at 12:12
1
What do you get by squaring both sides?
– Dietrich Burde
Dec 22 at 12:12
Just pay attention to the fact that when you substitute $;t:=sqrt x;$ , the solutions you take must be non-negative, as we have by agreement that $;sqrt xge0;$ .
– DonAntonio
Dec 22 at 12:27
But if it was a cubic, would it be ok to have a negative answer?
– Ma Ha
Dec 22 at 12:31
1
1
Substitute $t=sqrt{x}$
– Matko
Dec 22 at 12:12
Substitute $t=sqrt{x}$
– Matko
Dec 22 at 12:12
1
1
What do you get by squaring both sides?
– Dietrich Burde
Dec 22 at 12:12
What do you get by squaring both sides?
– Dietrich Burde
Dec 22 at 12:12
Just pay attention to the fact that when you substitute $;t:=sqrt x;$ , the solutions you take must be non-negative, as we have by agreement that $;sqrt xge0;$ .
– DonAntonio
Dec 22 at 12:27
Just pay attention to the fact that when you substitute $;t:=sqrt x;$ , the solutions you take must be non-negative, as we have by agreement that $;sqrt xge0;$ .
– DonAntonio
Dec 22 at 12:27
But if it was a cubic, would it be ok to have a negative answer?
– Ma Ha
Dec 22 at 12:31
But if it was a cubic, would it be ok to have a negative answer?
– Ma Ha
Dec 22 at 12:31
add a comment |
2 Answers
2
active
oldest
votes
Hint: change variable $y = sqrt{x}$ .
$$y + frac{8}{y} = 6 Rightarrow y^2- 6y + 8 = 0 Rightarrow (y-2)(y-4) = 0 Rightarrow y in {2, 4} Rightarrow sqrt{x} = 2, 4Rightarrow x in {4,16}$$
answered Dec 22 at 12:12
OmG
2,222620
add a comment |
Yor equation is equivalent to $$x+8=6sqrt{x}$$ with $$t=sqrt{x}$$ we get $$t^2-6t+8=0$$
answered Dec 22 at 12:16
Dr. Sonnhard Graubner
72.8k32865
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
Hint: change variable $y = sqrt{x}$ .
$$y + frac{8}{y} = 6 Rightarrow y^2- 6y + 8 = 0 Rightarrow (y-2)(y-4) = 0 Rightarrow y in {2, 4} Rightarrow sqrt{x} = 2, 4Rightarrow x in {4,16}$$
answered Dec 22 at 12:12
OmG
2,222620
add a comment |
Hint: change variable $y = sqrt{x}$ .
$$y + frac{8}{y} = 6 Rightarrow y^2- 6y + 8 = 0 Rightarrow (y-2)(y-4) = 0 Rightarrow y in {2, 4} Rightarrow sqrt{x} = 2, 4Rightarrow x in {4,16}$$
answered Dec 22 at 12:12
OmG
2,222620
add a comment |
Hint: change variable $y = sqrt{x}$ .
$$y + frac{8}{y} = 6 Rightarrow y^2- 6y + 8 = 0 Rightarrow (y-2)(y-4) = 0 Rightarrow y in {2, 4} Rightarrow sqrt{x} = 2, 4Rightarrow x in {4,16}$$
answered Dec 22 at 12:12
OmG
2,222620
Hint: change variable $y = sqrt{x}$ .
$$y + frac{8}{y} = 6 Rightarrow y^2- 6y + 8 = 0 Rightarrow (y-2)(y-4) = 0 Rightarrow y in {2, 4} Rightarrow sqrt{x} = 2, 4Rightarrow x in {4,16}$$
answered Dec 22 at 12:12
OmG
2,222620
answered Dec 22 at 12:12
OmG
2,222620
answered Dec 22 at 12:12
OmG
2,222620
answered Dec 22 at 12:12
OmG
2,222620
2,222620
add a comment |
add a comment |
Yor equation is equivalent to $$x+8=6sqrt{x}$$ with $$t=sqrt{x}$$ we get $$t^2-6t+8=0$$
answered Dec 22 at 12:16
Dr. Sonnhard Graubner
72.8k32865
add a comment |
Yor equation is equivalent to $$x+8=6sqrt{x}$$ with $$t=sqrt{x}$$ we get $$t^2-6t+8=0$$
answered Dec 22 at 12:16
Dr. Sonnhard Graubner
72.8k32865
add a comment |
Yor equation is equivalent to $$x+8=6sqrt{x}$$ with $$t=sqrt{x}$$ we get $$t^2-6t+8=0$$
answered Dec 22 at 12:16
Dr. Sonnhard Graubner
72.8k32865
Yor equation is equivalent to $$x+8=6sqrt{x}$$ with $$t=sqrt{x}$$ we get $$t^2-6t+8=0$$
answered Dec 22 at 12:16
Dr. Sonnhard Graubner
72.8k32865
answered Dec 22 at 12:16
Dr. Sonnhard Graubner
72.8k32865
answered Dec 22 at 12:16
Dr. Sonnhard Graubner
72.8k32865
answered Dec 22 at 12:16
Dr. Sonnhard Graubner
72.8k32865
72.8k32865
add a comment |
add a comment |
1
Substitute $t=sqrt{x}$
– Matko
Dec 22 at 12:12
1
What do you get by squaring both sides?
– Dietrich Burde
Dec 22 at 12:12
Just pay attention to the fact that when you substitute $;t:=sqrt x;$ , the solutions you take must be non-negative, as we have by agreement that $;sqrt xge0;$ .
– DonAntonio
Dec 22 at 12:27
But if it was a cubic, would it be ok to have a negative answer?
– Ma Ha
Dec 22 at 12:31