Find the exhaustive values of “a”, given the condition












2












$begingroup$



If the equation $2^{2x} + a*2^{x+1} + a + 1=0$ has roots of opposite sign then the exhaustive values of a are?




I tried taking $2^x = t$. But then didn't know what to do.



The equation became, $t^2 + 2at + a +1 =0$. But then, what conditions should I impose?



Since one root is negative and the other positive, the only conclusion that I could draw was that $x<0$ for the first condition and $x>0$ for the second condition. But, I don't know how do I proceed from here?



Any help would be appreciated.






exponential-function quadratics substitution







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  • $begingroup$
    In the original question, did you mean $2^{2x}+2^{x+1}a+a+1 = 0$?
    $endgroup$
    – KM101
    Jan 2 at 19:04












  • $begingroup$
    Yes! I am sorry for that funny looking character.
    $endgroup$
    – Tony
    Jan 2 at 19:05
















2












$begingroup$



If the equation $2^{2x} + a*2^{x+1} + a + 1=0$ has roots of opposite sign then the exhaustive values of a are?




I tried taking $2^x = t$. But then didn't know what to do.



The equation became, $t^2 + 2at + a +1 =0$. But then, what conditions should I impose?



Since one root is negative and the other positive, the only conclusion that I could draw was that $x<0$ for the first condition and $x>0$ for the second condition. But, I don't know how do I proceed from here?



Any help would be appreciated.






exponential-function quadratics substitution







share|cite|improve this question











$endgroup$












  • $begingroup$
    In the original question, did you mean $2^{2x}+2^{x+1}a+a+1 = 0$?
    $endgroup$
    – KM101
    Jan 2 at 19:04












  • $begingroup$
    Yes! I am sorry for that funny looking character.
    $endgroup$
    – Tony
    Jan 2 at 19:05














2












2








2





$begingroup$



If the equation $2^{2x} + a*2^{x+1} + a + 1=0$ has roots of opposite sign then the exhaustive values of a are?




I tried taking $2^x = t$. But then didn't know what to do.



The equation became, $t^2 + 2at + a +1 =0$. But then, what conditions should I impose?



Since one root is negative and the other positive, the only conclusion that I could draw was that $x<0$ for the first condition and $x>0$ for the second condition. But, I don't know how do I proceed from here?



Any help would be appreciated.






exponential-function quadratics substitution







share|cite|improve this question











$endgroup$





If the equation $2^{2x} + a*2^{x+1} + a + 1=0$ has roots of opposite sign then the exhaustive values of a are?




I tried taking $2^x = t$. But then didn't know what to do.



The equation became, $t^2 + 2at + a +1 =0$. But then, what conditions should I impose?



Since one root is negative and the other positive, the only conclusion that I could draw was that $x<0$ for the first condition and $x>0$ for the second condition. But, I don't know how do I proceed from here?



Any help would be appreciated.






exponential-function quadratics substitution




exponential-function quadratics substitution






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 2 at 19:13









Michael Rozenberg

107k1894198




107k1894198










asked Jan 2 at 19:00









TonyTony

523




523












  • $begingroup$
    In the original question, did you mean $2^{2x}+2^{x+1}a+a+1 = 0$?
    $endgroup$
    – KM101
    Jan 2 at 19:04












  • $begingroup$
    Yes! I am sorry for that funny looking character.
    $endgroup$
    – Tony
    Jan 2 at 19:05


















  • $begingroup$
    In the original question, did you mean $2^{2x}+2^{x+1}a+a+1 = 0$?
    $endgroup$
    – KM101
    Jan 2 at 19:04












  • $begingroup$
    Yes! I am sorry for that funny looking character.
    $endgroup$
    – Tony
    Jan 2 at 19:05
















$begingroup$
In the original question, did you mean $2^{2x}+2^{x+1}a+a+1 = 0$?
$endgroup$
– KM101
Jan 2 at 19:04






$begingroup$
In the original question, did you mean $2^{2x}+2^{x+1}a+a+1 = 0$?
$endgroup$
– KM101
Jan 2 at 19:04














$begingroup$
Yes! I am sorry for that funny looking character.
$endgroup$
– Tony
Jan 2 at 19:05




$begingroup$
Yes! I am sorry for that funny looking character.
$endgroup$
– Tony
Jan 2 at 19:05










3 Answers
3






active

oldest

votes


















2












$begingroup$

So far, so good! Now: if $x>0$, then $t=2^x>1$; and if $x<0$, then $t=2^x<1$. So for your new equation $t^2+2at+a+1=0$ you want it to have two roots $t$ of which one is greater and one is less than $1$. Comparing roots with $1$ seems a bit difficult; comparing with $0$ would be easier… So let's make another change of variables: $t$ is greater than or less than $1$ if and only if $y=t-1$ is greater than or less than $0$ respectively. Substituting $y=t-1$, i.e. $t=y+1$, the quadratic equations transforms into
$$begin{split}
t^2+2at+a+1=0 &iff (y+1)^2+2a(y+1)+a+1=0\
&iff y^2+2(a+1)y+(3a+2)=0.
end{split}$$



Now, we have to satisfy two conditions:




  • A quadratic equation (with real coefficients and the leading term of $1$, as is the case here) has roots of opposite signs if and only if its constant term is negative, so we need $3a+2<0$.

  • But also don't forget that $t=2^x$ can't be negative! In other words, $t>0$, and therefore $y=t-1>-1$. For a parabola $f(y)=y^2+2(a+1)y+(3a+2)$ opening up with one positive and one negative root, we need to guarantee that even the negative root is greater than $-1$, or in other words, that $-1$ lies to the left of both roots. This conditions will be satisfied if $f(-1)>0$, i.e. $(-1)^2+2(a+1)cdot(-1)+(3a+2)>0$, which simplifies to $a+1>0$.






share|cite|improve this answer











$endgroup$





















    2












    $begingroup$

    The hint:



    Solve the following system.
    $$1^2+2acdot1+a+1<0$$ and
    $$a+1>0.$$
    The first inequality says that $1$ is placed between $2^{x_1}$ and $2^{x_2}.$



    The second inequality says that the smaller root of the quadratic equation is positive.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Can you explain your steps a bit? Why do they make sense?
      $endgroup$
      – Tony
      Jan 2 at 19:25










    • $begingroup$
      @Sahil Baor I added and fixed something. See now.
      $endgroup$
      – Michael Rozenberg
      Jan 2 at 19:32










    • $begingroup$
      @Sahil Baori Because it's iff. The first inequality says that if $x_1<x_2$ then $x_2>0$ and $x_1<0$. The second inequality says that $x_1$ exists.
      $endgroup$
      – Michael Rozenberg
      Jan 2 at 19:43










    • $begingroup$
      @SahilBaori $f(t)=t^2+2at+a+1$ is an upward opening parabola with zeroes $t_1in(0,1),t_2>1$. This means $f(1)<0$. The product of roots $t_1t_2=a+1>0$, which, coupled with the first condition, implies $t_1,t_2>0$
      $endgroup$
      – Shubham Johri
      Jan 2 at 19:45





















    -1












    $begingroup$

    In order for distinct real roots, $a^2-a-1>0$.



    Since $x_1>0,2^{x_1}=t_1=-a+sqrt{a^2-a-1}>1$.



    Similarly, $x_2<0therefore 0<2^{x_2}=t_2=-a-sqrt{a^2-a-1}<1$.






    share|cite|improve this answer









    $endgroup$













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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      2












      $begingroup$

      So far, so good! Now: if $x>0$, then $t=2^x>1$; and if $x<0$, then $t=2^x<1$. So for your new equation $t^2+2at+a+1=0$ you want it to have two roots $t$ of which one is greater and one is less than $1$. Comparing roots with $1$ seems a bit difficult; comparing with $0$ would be easier… So let's make another change of variables: $t$ is greater than or less than $1$ if and only if $y=t-1$ is greater than or less than $0$ respectively. Substituting $y=t-1$, i.e. $t=y+1$, the quadratic equations transforms into
      $$begin{split}
      t^2+2at+a+1=0 &iff (y+1)^2+2a(y+1)+a+1=0\
      &iff y^2+2(a+1)y+(3a+2)=0.
      end{split}$$



      Now, we have to satisfy two conditions:




      • A quadratic equation (with real coefficients and the leading term of $1$, as is the case here) has roots of opposite signs if and only if its constant term is negative, so we need $3a+2<0$.

      • But also don't forget that $t=2^x$ can't be negative! In other words, $t>0$, and therefore $y=t-1>-1$. For a parabola $f(y)=y^2+2(a+1)y+(3a+2)$ opening up with one positive and one negative root, we need to guarantee that even the negative root is greater than $-1$, or in other words, that $-1$ lies to the left of both roots. This conditions will be satisfied if $f(-1)>0$, i.e. $(-1)^2+2(a+1)cdot(-1)+(3a+2)>0$, which simplifies to $a+1>0$.






      share|cite|improve this answer











      $endgroup$


















        2












        $begingroup$

        So far, so good! Now: if $x>0$, then $t=2^x>1$; and if $x<0$, then $t=2^x<1$. So for your new equation $t^2+2at+a+1=0$ you want it to have two roots $t$ of which one is greater and one is less than $1$. Comparing roots with $1$ seems a bit difficult; comparing with $0$ would be easier… So let's make another change of variables: $t$ is greater than or less than $1$ if and only if $y=t-1$ is greater than or less than $0$ respectively. Substituting $y=t-1$, i.e. $t=y+1$, the quadratic equations transforms into
        $$begin{split}
        t^2+2at+a+1=0 &iff (y+1)^2+2a(y+1)+a+1=0\
        &iff y^2+2(a+1)y+(3a+2)=0.
        end{split}$$



        Now, we have to satisfy two conditions:




        • A quadratic equation (with real coefficients and the leading term of $1$, as is the case here) has roots of opposite signs if and only if its constant term is negative, so we need $3a+2<0$.

        • But also don't forget that $t=2^x$ can't be negative! In other words, $t>0$, and therefore $y=t-1>-1$. For a parabola $f(y)=y^2+2(a+1)y+(3a+2)$ opening up with one positive and one negative root, we need to guarantee that even the negative root is greater than $-1$, or in other words, that $-1$ lies to the left of both roots. This conditions will be satisfied if $f(-1)>0$, i.e. $(-1)^2+2(a+1)cdot(-1)+(3a+2)>0$, which simplifies to $a+1>0$.






        share|cite|improve this answer











        $endgroup$
















          2












          2








          2





          $begingroup$

          So far, so good! Now: if $x>0$, then $t=2^x>1$; and if $x<0$, then $t=2^x<1$. So for your new equation $t^2+2at+a+1=0$ you want it to have two roots $t$ of which one is greater and one is less than $1$. Comparing roots with $1$ seems a bit difficult; comparing with $0$ would be easier… So let's make another change of variables: $t$ is greater than or less than $1$ if and only if $y=t-1$ is greater than or less than $0$ respectively. Substituting $y=t-1$, i.e. $t=y+1$, the quadratic equations transforms into
          $$begin{split}
          t^2+2at+a+1=0 &iff (y+1)^2+2a(y+1)+a+1=0\
          &iff y^2+2(a+1)y+(3a+2)=0.
          end{split}$$



          Now, we have to satisfy two conditions:




          • A quadratic equation (with real coefficients and the leading term of $1$, as is the case here) has roots of opposite signs if and only if its constant term is negative, so we need $3a+2<0$.

          • But also don't forget that $t=2^x$ can't be negative! In other words, $t>0$, and therefore $y=t-1>-1$. For a parabola $f(y)=y^2+2(a+1)y+(3a+2)$ opening up with one positive and one negative root, we need to guarantee that even the negative root is greater than $-1$, or in other words, that $-1$ lies to the left of both roots. This conditions will be satisfied if $f(-1)>0$, i.e. $(-1)^2+2(a+1)cdot(-1)+(3a+2)>0$, which simplifies to $a+1>0$.






          share|cite|improve this answer











          $endgroup$



          So far, so good! Now: if $x>0$, then $t=2^x>1$; and if $x<0$, then $t=2^x<1$. So for your new equation $t^2+2at+a+1=0$ you want it to have two roots $t$ of which one is greater and one is less than $1$. Comparing roots with $1$ seems a bit difficult; comparing with $0$ would be easier… So let's make another change of variables: $t$ is greater than or less than $1$ if and only if $y=t-1$ is greater than or less than $0$ respectively. Substituting $y=t-1$, i.e. $t=y+1$, the quadratic equations transforms into
          $$begin{split}
          t^2+2at+a+1=0 &iff (y+1)^2+2a(y+1)+a+1=0\
          &iff y^2+2(a+1)y+(3a+2)=0.
          end{split}$$



          Now, we have to satisfy two conditions:




          • A quadratic equation (with real coefficients and the leading term of $1$, as is the case here) has roots of opposite signs if and only if its constant term is negative, so we need $3a+2<0$.

          • But also don't forget that $t=2^x$ can't be negative! In other words, $t>0$, and therefore $y=t-1>-1$. For a parabola $f(y)=y^2+2(a+1)y+(3a+2)$ opening up with one positive and one negative root, we need to guarantee that even the negative root is greater than $-1$, or in other words, that $-1$ lies to the left of both roots. This conditions will be satisfied if $f(-1)>0$, i.e. $(-1)^2+2(a+1)cdot(-1)+(3a+2)>0$, which simplifies to $a+1>0$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 2 at 20:05

























          answered Jan 2 at 19:32









          zipirovichzipirovich

          11.3k11731




          11.3k11731























              2












              $begingroup$

              The hint:



              Solve the following system.
              $$1^2+2acdot1+a+1<0$$ and
              $$a+1>0.$$
              The first inequality says that $1$ is placed between $2^{x_1}$ and $2^{x_2}.$



              The second inequality says that the smaller root of the quadratic equation is positive.






              share|cite|improve this answer











              $endgroup$













              • $begingroup$
                Can you explain your steps a bit? Why do they make sense?
                $endgroup$
                – Tony
                Jan 2 at 19:25










              • $begingroup$
                @Sahil Baor I added and fixed something. See now.
                $endgroup$
                – Michael Rozenberg
                Jan 2 at 19:32










              • $begingroup$
                @Sahil Baori Because it's iff. The first inequality says that if $x_1<x_2$ then $x_2>0$ and $x_1<0$. The second inequality says that $x_1$ exists.
                $endgroup$
                – Michael Rozenberg
                Jan 2 at 19:43










              • $begingroup$
                @SahilBaori $f(t)=t^2+2at+a+1$ is an upward opening parabola with zeroes $t_1in(0,1),t_2>1$. This means $f(1)<0$. The product of roots $t_1t_2=a+1>0$, which, coupled with the first condition, implies $t_1,t_2>0$
                $endgroup$
                – Shubham Johri
                Jan 2 at 19:45


















              2












              $begingroup$

              The hint:



              Solve the following system.
              $$1^2+2acdot1+a+1<0$$ and
              $$a+1>0.$$
              The first inequality says that $1$ is placed between $2^{x_1}$ and $2^{x_2}.$



              The second inequality says that the smaller root of the quadratic equation is positive.






              share|cite|improve this answer











              $endgroup$













              • $begingroup$
                Can you explain your steps a bit? Why do they make sense?
                $endgroup$
                – Tony
                Jan 2 at 19:25










              • $begingroup$
                @Sahil Baor I added and fixed something. See now.
                $endgroup$
                – Michael Rozenberg
                Jan 2 at 19:32










              • $begingroup$
                @Sahil Baori Because it's iff. The first inequality says that if $x_1<x_2$ then $x_2>0$ and $x_1<0$. The second inequality says that $x_1$ exists.
                $endgroup$
                – Michael Rozenberg
                Jan 2 at 19:43










              • $begingroup$
                @SahilBaori $f(t)=t^2+2at+a+1$ is an upward opening parabola with zeroes $t_1in(0,1),t_2>1$. This means $f(1)<0$. The product of roots $t_1t_2=a+1>0$, which, coupled with the first condition, implies $t_1,t_2>0$
                $endgroup$
                – Shubham Johri
                Jan 2 at 19:45
















              2












              2








              2





              $begingroup$

              The hint:



              Solve the following system.
              $$1^2+2acdot1+a+1<0$$ and
              $$a+1>0.$$
              The first inequality says that $1$ is placed between $2^{x_1}$ and $2^{x_2}.$



              The second inequality says that the smaller root of the quadratic equation is positive.






              share|cite|improve this answer











              $endgroup$



              The hint:



              Solve the following system.
              $$1^2+2acdot1+a+1<0$$ and
              $$a+1>0.$$
              The first inequality says that $1$ is placed between $2^{x_1}$ and $2^{x_2}.$



              The second inequality says that the smaller root of the quadratic equation is positive.







              share|cite|improve this answer














              share|cite|improve this answer



              share|cite|improve this answer








              edited Jan 2 at 19:30

























              answered Jan 2 at 19:10









              Michael RozenbergMichael Rozenberg

              107k1894198




              107k1894198












              • $begingroup$
                Can you explain your steps a bit? Why do they make sense?
                $endgroup$
                – Tony
                Jan 2 at 19:25










              • $begingroup$
                @Sahil Baor I added and fixed something. See now.
                $endgroup$
                – Michael Rozenberg
                Jan 2 at 19:32










              • $begingroup$
                @Sahil Baori Because it's iff. The first inequality says that if $x_1<x_2$ then $x_2>0$ and $x_1<0$. The second inequality says that $x_1$ exists.
                $endgroup$
                – Michael Rozenberg
                Jan 2 at 19:43










              • $begingroup$
                @SahilBaori $f(t)=t^2+2at+a+1$ is an upward opening parabola with zeroes $t_1in(0,1),t_2>1$. This means $f(1)<0$. The product of roots $t_1t_2=a+1>0$, which, coupled with the first condition, implies $t_1,t_2>0$
                $endgroup$
                – Shubham Johri
                Jan 2 at 19:45




















              • $begingroup$
                Can you explain your steps a bit? Why do they make sense?
                $endgroup$
                – Tony
                Jan 2 at 19:25










              • $begingroup$
                @Sahil Baor I added and fixed something. See now.
                $endgroup$
                – Michael Rozenberg
                Jan 2 at 19:32










              • $begingroup$
                @Sahil Baori Because it's iff. The first inequality says that if $x_1<x_2$ then $x_2>0$ and $x_1<0$. The second inequality says that $x_1$ exists.
                $endgroup$
                – Michael Rozenberg
                Jan 2 at 19:43










              • $begingroup$
                @SahilBaori $f(t)=t^2+2at+a+1$ is an upward opening parabola with zeroes $t_1in(0,1),t_2>1$. This means $f(1)<0$. The product of roots $t_1t_2=a+1>0$, which, coupled with the first condition, implies $t_1,t_2>0$
                $endgroup$
                – Shubham Johri
                Jan 2 at 19:45


















              $begingroup$
              Can you explain your steps a bit? Why do they make sense?
              $endgroup$
              – Tony
              Jan 2 at 19:25




              $begingroup$
              Can you explain your steps a bit? Why do they make sense?
              $endgroup$
              – Tony
              Jan 2 at 19:25












              $begingroup$
              @Sahil Baor I added and fixed something. See now.
              $endgroup$
              – Michael Rozenberg
              Jan 2 at 19:32




              $begingroup$
              @Sahil Baor I added and fixed something. See now.
              $endgroup$
              – Michael Rozenberg
              Jan 2 at 19:32












              $begingroup$
              @Sahil Baori Because it's iff. The first inequality says that if $x_1<x_2$ then $x_2>0$ and $x_1<0$. The second inequality says that $x_1$ exists.
              $endgroup$
              – Michael Rozenberg
              Jan 2 at 19:43




              $begingroup$
              @Sahil Baori Because it's iff. The first inequality says that if $x_1<x_2$ then $x_2>0$ and $x_1<0$. The second inequality says that $x_1$ exists.
              $endgroup$
              – Michael Rozenberg
              Jan 2 at 19:43












              $begingroup$
              @SahilBaori $f(t)=t^2+2at+a+1$ is an upward opening parabola with zeroes $t_1in(0,1),t_2>1$. This means $f(1)<0$. The product of roots $t_1t_2=a+1>0$, which, coupled with the first condition, implies $t_1,t_2>0$
              $endgroup$
              – Shubham Johri
              Jan 2 at 19:45






              $begingroup$
              @SahilBaori $f(t)=t^2+2at+a+1$ is an upward opening parabola with zeroes $t_1in(0,1),t_2>1$. This means $f(1)<0$. The product of roots $t_1t_2=a+1>0$, which, coupled with the first condition, implies $t_1,t_2>0$
              $endgroup$
              – Shubham Johri
              Jan 2 at 19:45













              -1












              $begingroup$

              In order for distinct real roots, $a^2-a-1>0$.



              Since $x_1>0,2^{x_1}=t_1=-a+sqrt{a^2-a-1}>1$.



              Similarly, $x_2<0therefore 0<2^{x_2}=t_2=-a-sqrt{a^2-a-1}<1$.






              share|cite|improve this answer









              $endgroup$


















                -1












                $begingroup$

                In order for distinct real roots, $a^2-a-1>0$.



                Since $x_1>0,2^{x_1}=t_1=-a+sqrt{a^2-a-1}>1$.



                Similarly, $x_2<0therefore 0<2^{x_2}=t_2=-a-sqrt{a^2-a-1}<1$.






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                $endgroup$
















                  -1












                  -1








                  -1





                  $begingroup$

                  In order for distinct real roots, $a^2-a-1>0$.



                  Since $x_1>0,2^{x_1}=t_1=-a+sqrt{a^2-a-1}>1$.



                  Similarly, $x_2<0therefore 0<2^{x_2}=t_2=-a-sqrt{a^2-a-1}<1$.






                  share|cite|improve this answer









                  $endgroup$



                  In order for distinct real roots, $a^2-a-1>0$.



                  Since $x_1>0,2^{x_1}=t_1=-a+sqrt{a^2-a-1}>1$.



                  Similarly, $x_2<0therefore 0<2^{x_2}=t_2=-a-sqrt{a^2-a-1}<1$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 2 at 19:26









                  Shubham JohriShubham Johri

                  5,204718




                  5,204718






























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