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.



  1. the discriminant is $D=(k-2sqrt2)(k+2sqrt2)$.

  2. the quadratic equation gives $3^x=dfrac-kpmsqrtk^2-82$.

  3. 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




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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














up vote
1
down vote

favorite












I have worked this one through but still not 100% sure.



  1. the discriminant is $D=(k-2sqrt2)(k+2sqrt2)$.

  2. the quadratic equation gives $3^x=dfrac-kpmsqrtk^2-82$.

  3. 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




share|cite|improve this question
















  • 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












up vote
1
down vote

favorite









up vote
1
down vote

favorite











I have worked this one through but still not 100% sure.



  1. the discriminant is $D=(k-2sqrt2)(k+2sqrt2)$.

  2. the quadratic equation gives $3^x=dfrac-kpmsqrtk^2-82$.

  3. 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




share|cite|improve this question












I have worked this one through but still not 100% sure.



  1. the discriminant is $D=(k-2sqrt2)(k+2sqrt2)$.

  2. the quadratic equation gives $3^x=dfrac-kpmsqrtk^2-82$.

  3. 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





share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










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












  • 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










1 Answer
1






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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.



enter image description here






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  • Thank you. This makes sense now.
    – Will Kim
    Aug 28 at 15:28










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1 Answer
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active

oldest

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1 Answer
1






active

oldest

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active

oldest

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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.



enter image description here






share|cite|improve this answer






















  • Thank you. This makes sense now.
    – Will Kim
    Aug 28 at 15:28














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.



enter image description here






share|cite|improve this answer






















  • Thank you. This makes sense now.
    – Will Kim
    Aug 28 at 15:28












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.



enter image description here






share|cite|improve this answer














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.



enter image description here







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Aug 28 at 14:41

























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
















  • 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

















 

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