Number of solution of $x^4-5x^3+(lambda+2)x^2-5x+1=0$

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Consider the bi-quadratic equation $E:x^4-5x^3+(lambda+2)x^2-5x+1=0$ then, the real values of $lambda$ so that $E$ has four different solutions is?




My attempts:



As $x=0$ is not a solution for any $lambda$ hence we divide by $x^2$,beginalignx^2+dfrac1x^2-5x-5dfrac1x+lambda+2&=0\bigg(x+dfrac1xbigg)^2-5bigg(x+dfrac1xbigg)&=-lambda\t(t-5)&=-lambdaendalign



Here, $x+dfrac1x=t$ cannot lie between $[-2,2]$.



Now we draw the graph of $-t(t-5)$ and $lambda$ if they intersect twice then $t_1,t_2$ are the roots of $Q:t^2-5t+lambda=0$, each of the roots of $Q$ gives two $x's$.



From GRAPH HERE, my solution is $lambdain(-infty,-14]cup[6,frac254)$



But this wrong according to answer provided, please help.




algebra-precalculus polynomials roots graphing-functions quartic-equations




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  • 3




    This is not a biquadratic equation, but a reciprocal quartic equation.
    – Bernard
    Aug 1 at 13:21














up vote
2
down vote

favorite













Consider the bi-quadratic equation $E:x^4-5x^3+(lambda+2)x^2-5x+1=0$ then, the real values of $lambda$ so that $E$ has four different solutions is?




My attempts:



As $x=0$ is not a solution for any $lambda$ hence we divide by $x^2$,beginalignx^2+dfrac1x^2-5x-5dfrac1x+lambda+2&=0\bigg(x+dfrac1xbigg)^2-5bigg(x+dfrac1xbigg)&=-lambda\t(t-5)&=-lambdaendalign



Here, $x+dfrac1x=t$ cannot lie between $[-2,2]$.



Now we draw the graph of $-t(t-5)$ and $lambda$ if they intersect twice then $t_1,t_2$ are the roots of $Q:t^2-5t+lambda=0$, each of the roots of $Q$ gives two $x's$.



From GRAPH HERE, my solution is $lambdain(-infty,-14]cup[6,frac254)$



But this wrong according to answer provided, please help.




algebra-precalculus polynomials roots graphing-functions quartic-equations




share|cite|improve this question

















  • 3




    This is not a biquadratic equation, but a reciprocal quartic equation.
    – Bernard
    Aug 1 at 13:21












up vote
2
down vote

favorite









up vote
2
down vote

favorite












Consider the bi-quadratic equation $E:x^4-5x^3+(lambda+2)x^2-5x+1=0$ then, the real values of $lambda$ so that $E$ has four different solutions is?




My attempts:



As $x=0$ is not a solution for any $lambda$ hence we divide by $x^2$,beginalignx^2+dfrac1x^2-5x-5dfrac1x+lambda+2&=0\bigg(x+dfrac1xbigg)^2-5bigg(x+dfrac1xbigg)&=-lambda\t(t-5)&=-lambdaendalign



Here, $x+dfrac1x=t$ cannot lie between $[-2,2]$.



Now we draw the graph of $-t(t-5)$ and $lambda$ if they intersect twice then $t_1,t_2$ are the roots of $Q:t^2-5t+lambda=0$, each of the roots of $Q$ gives two $x's$.



From GRAPH HERE, my solution is $lambdain(-infty,-14]cup[6,frac254)$



But this wrong according to answer provided, please help.




algebra-precalculus polynomials roots graphing-functions quartic-equations




share|cite|improve this question














Consider the bi-quadratic equation $E:x^4-5x^3+(lambda+2)x^2-5x+1=0$ then, the real values of $lambda$ so that $E$ has four different solutions is?




My attempts:



As $x=0$ is not a solution for any $lambda$ hence we divide by $x^2$,beginalignx^2+dfrac1x^2-5x-5dfrac1x+lambda+2&=0\bigg(x+dfrac1xbigg)^2-5bigg(x+dfrac1xbigg)&=-lambda\t(t-5)&=-lambdaendalign



Here, $x+dfrac1x=t$ cannot lie between $[-2,2]$.



Now we draw the graph of $-t(t-5)$ and $lambda$ if they intersect twice then $t_1,t_2$ are the roots of $Q:t^2-5t+lambda=0$, each of the roots of $Q$ gives two $x's$.



From GRAPH HERE, my solution is $lambdain(-infty,-14]cup[6,frac254)$



But this wrong according to answer provided, please help.





algebra-precalculus polynomials roots graphing-functions quartic-equations





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edited Aug 1 at 14:03









Michael Rozenberg

87.4k1577179




87.4k1577179









asked Aug 1 at 13:06









mnulb

1,311620




1,311620







  • 3




    This is not a biquadratic equation, but a reciprocal quartic equation.
    – Bernard
    Aug 1 at 13:21












  • 3




    This is not a biquadratic equation, but a reciprocal quartic equation.
    – Bernard
    Aug 1 at 13:21







3




3




This is not a biquadratic equation, but a reciprocal quartic equation.
– Bernard
Aug 1 at 13:21




This is not a biquadratic equation, but a reciprocal quartic equation.
– Bernard
Aug 1 at 13:21










2 Answers
2






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Let $f(x)=-fracx^4-5x^3+2x^2-5x+1x^2.$



We need to find all values of $lambda,$ for which the equation
$$lambda=f(x)$$ has four different roots.



Now, $$f'(x)=frac(2-x)(x-1)(2x-1)(x+1)x^3,$$ which says that
$$lambda<f(-1)$$ or $$f(1)<lambda<minleftfleft(frac12right),f(2)right,$$
which gives the answer:
$$(-infty,-14)cup(6,6.25).$$






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    Hint: Draw a graph $$f(x)= -x^4+5x^3-2x^2+5x-1over x^2$$
    and see for which $lambda$ the line $y= lambda$ cuts the graph of $f$ four times.






    share|cite|improve this answer





















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      2 Answers
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      2 Answers
      2






      active

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      active

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      up vote
      2
      down vote













      Let $f(x)=-fracx^4-5x^3+2x^2-5x+1x^2.$



      We need to find all values of $lambda,$ for which the equation
      $$lambda=f(x)$$ has four different roots.



      Now, $$f'(x)=frac(2-x)(x-1)(2x-1)(x+1)x^3,$$ which says that
      $$lambda<f(-1)$$ or $$f(1)<lambda<minleftfleft(frac12right),f(2)right,$$
      which gives the answer:
      $$(-infty,-14)cup(6,6.25).$$






      share|cite|improve this answer



























        up vote
        2
        down vote













        Let $f(x)=-fracx^4-5x^3+2x^2-5x+1x^2.$



        We need to find all values of $lambda,$ for which the equation
        $$lambda=f(x)$$ has four different roots.



        Now, $$f'(x)=frac(2-x)(x-1)(2x-1)(x+1)x^3,$$ which says that
        $$lambda<f(-1)$$ or $$f(1)<lambda<minleftfleft(frac12right),f(2)right,$$
        which gives the answer:
        $$(-infty,-14)cup(6,6.25).$$






        share|cite|improve this answer

























          up vote
          2
          down vote










          up vote
          2
          down vote









          Let $f(x)=-fracx^4-5x^3+2x^2-5x+1x^2.$



          We need to find all values of $lambda,$ for which the equation
          $$lambda=f(x)$$ has four different roots.



          Now, $$f'(x)=frac(2-x)(x-1)(2x-1)(x+1)x^3,$$ which says that
          $$lambda<f(-1)$$ or $$f(1)<lambda<minleftfleft(frac12right),f(2)right,$$
          which gives the answer:
          $$(-infty,-14)cup(6,6.25).$$






          share|cite|improve this answer















          Let $f(x)=-fracx^4-5x^3+2x^2-5x+1x^2.$



          We need to find all values of $lambda,$ for which the equation
          $$lambda=f(x)$$ has four different roots.



          Now, $$f'(x)=frac(2-x)(x-1)(2x-1)(x+1)x^3,$$ which says that
          $$lambda<f(-1)$$ or $$f(1)<lambda<minleftfleft(frac12right),f(2)right,$$
          which gives the answer:
          $$(-infty,-14)cup(6,6.25).$$







          share|cite|improve this answer















          share|cite|improve this answer



          share|cite|improve this answer








          edited Aug 1 at 13:37


























          answered Aug 1 at 13:32









          Michael Rozenberg

          87.4k1577179




          87.4k1577179




















              up vote
              0
              down vote













              Hint: Draw a graph $$f(x)= -x^4+5x^3-2x^2+5x-1over x^2$$
              and see for which $lambda$ the line $y= lambda$ cuts the graph of $f$ four times.






              share|cite|improve this answer

























                up vote
                0
                down vote













                Hint: Draw a graph $$f(x)= -x^4+5x^3-2x^2+5x-1over x^2$$
                and see for which $lambda$ the line $y= lambda$ cuts the graph of $f$ four times.






                share|cite|improve this answer























                  up vote
                  0
                  down vote










                  up vote
                  0
                  down vote









                  Hint: Draw a graph $$f(x)= -x^4+5x^3-2x^2+5x-1over x^2$$
                  and see for which $lambda$ the line $y= lambda$ cuts the graph of $f$ four times.






                  share|cite|improve this answer













                  Hint: Draw a graph $$f(x)= -x^4+5x^3-2x^2+5x-1over x^2$$
                  and see for which $lambda$ the line $y= lambda$ cuts the graph of $f$ four times.







                  share|cite|improve this answer













                  share|cite|improve this answer



                  share|cite|improve this answer











                  answered Aug 1 at 13:31









                  greedoid

                  26.1k93473




                  26.1k93473






















                       

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