Investigating the solutions of $9^x+k(3^x)+2=0$
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I have worked this one through but still not 100% sure.
- the discriminant is $D=(k-2sqrt2)(k+2sqrt2)$.
- the quadratic equation gives $3^x=dfrac-kpmsqrtk^2-82$.
- as the RHS must be at least $0$ for this equation to have any solutions I did some work using inequalities
I concluded that this equation:
cannot have repeated roots when $k=-2sqrt2$
has two distinct roots when $k<-2sqrt2$
But not too sure about the details for the case when the equation has two distinct roots. How would you determine this?
Thank you
discriminant
asked Aug 28 at 13:46
Will Kim
1345
add a comment |Â
up vote
1
down vote
favorite
I have worked this one through but still not 100% sure.
- the discriminant is $D=(k-2sqrt2)(k+2sqrt2)$.
- the quadratic equation gives $3^x=dfrac-kpmsqrtk^2-82$.
- as the RHS must be at least $0$ for this equation to have any solutions I did some work using inequalities
I concluded that this equation:
cannot have repeated roots when $k=-2sqrt2$
has two distinct roots when $k<-2sqrt2$
But not too sure about the details for the case when the equation has two distinct roots. How would you determine this?
Thank you
discriminant
asked Aug 28 at 13:46
Will Kim
1345
1
yep, you can also check using wolfram...wolframalpha.com/input/?i=9%5Ex%2B8%5E(1%2F2)(3%5Ex)%2B2+%3D+0
– Andrew Allen
Aug 28 at 13:52
1
you must have $3^x > 0$ if you are to finally get an $x$ out of it.
– Alvin Lepik
Aug 28 at 13:52
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I have worked this one through but still not 100% sure.
- the discriminant is $D=(k-2sqrt2)(k+2sqrt2)$.
- the quadratic equation gives $3^x=dfrac-kpmsqrtk^2-82$.
- as the RHS must be at least $0$ for this equation to have any solutions I did some work using inequalities
I concluded that this equation:
cannot have repeated roots when $k=-2sqrt2$
has two distinct roots when $k<-2sqrt2$
But not too sure about the details for the case when the equation has two distinct roots. How would you determine this?
Thank you
discriminant
asked Aug 28 at 13:46
Will Kim
1345
I have worked this one through but still not 100% sure.
- the discriminant is $D=(k-2sqrt2)(k+2sqrt2)$.
- the quadratic equation gives $3^x=dfrac-kpmsqrtk^2-82$.
- as the RHS must be at least $0$ for this equation to have any solutions I did some work using inequalities
I concluded that this equation:
cannot have repeated roots when $k=-2sqrt2$
has two distinct roots when $k<-2sqrt2$
But not too sure about the details for the case when the equation has two distinct roots. How would you determine this?
Thank you
discriminant
asked Aug 28 at 13:46
Will Kim
1345
asked Aug 28 at 13:46
Will Kim
1345
asked Aug 28 at 13:46
Will Kim
1345
asked Aug 28 at 13:46
Will Kim
1345
1345
1
yep, you can also check using wolfram...wolframalpha.com/input/?i=9%5Ex%2B8%5E(1%2F2)(3%5Ex)%2B2+%3D+0
– Andrew Allen
Aug 28 at 13:52
1
you must have $3^x > 0$ if you are to finally get an $x$ out of it.
– Alvin Lepik
Aug 28 at 13:52
add a comment |Â
1
yep, you can also check using wolfram...wolframalpha.com/input/?i=9%5Ex%2B8%5E(1%2F2)(3%5Ex)%2B2+%3D+0
– Andrew Allen
Aug 28 at 13:52
1
you must have $3^x > 0$ if you are to finally get an $x$ out of it.
– Alvin Lepik
Aug 28 at 13:52
1
1
yep, you can also check using wolfram...wolframalpha.com/input/?i=9%5Ex%2B8%5E(1%2F2)(3%5Ex)%2B2+%3D+0
– Andrew Allen
Aug 28 at 13:52
yep, you can also check using wolfram...wolframalpha.com/input/?i=9%5Ex%2B8%5E(1%2F2)(3%5Ex)%2B2+%3D+0
– Andrew Allen
Aug 28 at 13:52
1
1
you must have $3^x > 0$ if you are to finally get an $x$ out of it.
– Alvin Lepik
Aug 28 at 13:52
you must have $3^x > 0$ if you are to finally get an $x$ out of it.
– Alvin Lepik
Aug 28 at 13:52
add a comment |Â
1 Answer
1
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oldest
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up vote
2
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accepted
To be able to compute the roots of $3^2x + k3^x +2$ with respect to $3^x$ you have immediately the condition $k^2 - 8 geq 0$. So $kleq -2sqrt2$ or $kgeq 2sqrt2$. In case of either equality, you have zero discriminant.
You must also meet the condition
$$-k pm sqrtk^2-8 > 0. $$
This leaves you with $k < -2sqrt2$.
Solving analytically the inequality $-k pm sqrtk^2-8 > 0$ can be prone to mistakes so make sure to double check.
When in doubt, I can always use my rudimentary geogebra skills to draw a picture and get a feeling for what's going on. The green line is $f(x) = -x$ and blue and red lines are $g(x) = sqrtx^2-8$ and $h(x) = -g(x)$ respectively.
answered Aug 28 at 13:59
Alvin Lepik
2,528921
Thank you. This makes sense now.
– Will Kim
Aug 28 at 15:28
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
To be able to compute the roots of $3^2x + k3^x +2$ with respect to $3^x$ you have immediately the condition $k^2 - 8 geq 0$. So $kleq -2sqrt2$ or $kgeq 2sqrt2$. In case of either equality, you have zero discriminant.
You must also meet the condition
$$-k pm sqrtk^2-8 > 0. $$
This leaves you with $k < -2sqrt2$.
Solving analytically the inequality $-k pm sqrtk^2-8 > 0$ can be prone to mistakes so make sure to double check.
When in doubt, I can always use my rudimentary geogebra skills to draw a picture and get a feeling for what's going on. The green line is $f(x) = -x$ and blue and red lines are $g(x) = sqrtx^2-8$ and $h(x) = -g(x)$ respectively.
answered Aug 28 at 13:59
Alvin Lepik
2,528921
Thank you. This makes sense now.
– Will Kim
Aug 28 at 15:28
add a comment |Â
up vote
2
down vote
accepted
To be able to compute the roots of $3^2x + k3^x +2$ with respect to $3^x$ you have immediately the condition $k^2 - 8 geq 0$. So $kleq -2sqrt2$ or $kgeq 2sqrt2$. In case of either equality, you have zero discriminant.
You must also meet the condition
$$-k pm sqrtk^2-8 > 0. $$
This leaves you with $k < -2sqrt2$.
Solving analytically the inequality $-k pm sqrtk^2-8 > 0$ can be prone to mistakes so make sure to double check.
When in doubt, I can always use my rudimentary geogebra skills to draw a picture and get a feeling for what's going on. The green line is $f(x) = -x$ and blue and red lines are $g(x) = sqrtx^2-8$ and $h(x) = -g(x)$ respectively.
answered Aug 28 at 13:59
Alvin Lepik
2,528921
Thank you. This makes sense now.
– Will Kim
Aug 28 at 15:28
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
To be able to compute the roots of $3^2x + k3^x +2$ with respect to $3^x$ you have immediately the condition $k^2 - 8 geq 0$. So $kleq -2sqrt2$ or $kgeq 2sqrt2$. In case of either equality, you have zero discriminant.
You must also meet the condition
$$-k pm sqrtk^2-8 > 0. $$
This leaves you with $k < -2sqrt2$.
Solving analytically the inequality $-k pm sqrtk^2-8 > 0$ can be prone to mistakes so make sure to double check.
When in doubt, I can always use my rudimentary geogebra skills to draw a picture and get a feeling for what's going on. The green line is $f(x) = -x$ and blue and red lines are $g(x) = sqrtx^2-8$ and $h(x) = -g(x)$ respectively.
answered Aug 28 at 13:59
Alvin Lepik
2,528921
To be able to compute the roots of $3^2x + k3^x +2$ with respect to $3^x$ you have immediately the condition $k^2 - 8 geq 0$. So $kleq -2sqrt2$ or $kgeq 2sqrt2$. In case of either equality, you have zero discriminant.
You must also meet the condition
$$-k pm sqrtk^2-8 > 0. $$
This leaves you with $k < -2sqrt2$.
Solving analytically the inequality $-k pm sqrtk^2-8 > 0$ can be prone to mistakes so make sure to double check.
When in doubt, I can always use my rudimentary geogebra skills to draw a picture and get a feeling for what's going on. The green line is $f(x) = -x$ and blue and red lines are $g(x) = sqrtx^2-8$ and $h(x) = -g(x)$ respectively.
answered Aug 28 at 13:59
Alvin Lepik
2,528921
edited Aug 28 at 14:41
answered Aug 28 at 13:59
Alvin Lepik
2,528921
answered Aug 28 at 13:59
Alvin Lepik
2,528921
answered Aug 28 at 13:59
Alvin Lepik
2,528921
2,528921
Thank you. This makes sense now.
– Will Kim
Aug 28 at 15:28
add a comment |Â
Thank you. This makes sense now.
– Will Kim
Aug 28 at 15:28
Thank you. This makes sense now.
– Will Kim
Aug 28 at 15:28
Thank you. This makes sense now.
– Will Kim
Aug 28 at 15:28
add a comment |Â
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1
yep, you can also check using wolfram...wolframalpha.com/input/?i=9%5Ex%2B8%5E(1%2F2)(3%5Ex)%2B2+%3D+0
– Andrew Allen
Aug 28 at 13:52
1
you must have $3^x > 0$ if you are to finally get an $x$ out of it.
– Alvin Lepik
Aug 28 at 13:52