Find minimum value of $a^2+b^2$












4












$begingroup$


Given that $a$ and $b$ are real constants and that the equation $x^4+ax^3+2x^2+bx+1=0$ has at least one real root, find the minimum possible value of $a^2+b^2$.



I began this way: Let the polynomial be factorized as $(x^2+alpha x + 1)(x^2+beta x +1)$. Then expanding and comparing coefficients we get $alphabeta=0$, meaning either $alpha=0$ or $beta=0$. Suppose $alpha=0$. Then we see that $(x^2+beta x+1)$ should have real roots, from which we get $beta^2 geq 4$. But we get $a=b=beta$ from the comparison above. So $a^2+b^2 = 2beta^2 geq 8$.



Is it correct? Or is there any mistake? Any other solution is also welcome.






polynomials







share|cite|improve this question









$endgroup$








  • 2




    $begingroup$
    How do you know the polynomial has such a factorisation?
    $endgroup$
    – Lord Shark the Unknown
    Jan 10 at 5:27










  • $begingroup$
    Any fourth degree polynomial can be factorised into product of two quadratic expressions, no matter what the nature of its roots are, right?
    $endgroup$
    – Yellow
    Jan 10 at 5:32










  • $begingroup$
    @Any Radha, yes but not in that way you write...
    $endgroup$
    – dmtri
    Jan 10 at 5:38










  • $begingroup$
    I can't find anything wrong in your solution.
    $endgroup$
    – Anik Bhowmick
    Jan 10 at 5:43






  • 1




    $begingroup$
    Wouldn't this imply that $alpha + beta = a = b$ which need not be the case? If $x^4 + ax^3 + 2x^2 +bx +1 = (m x^2 + nx + p)(jx^2 + kx + q)$ we know $j = frac 1m$ and wolog it factors to $(x^2 + alpha x + gamma)(x^2 + beta x + delta)$ but we can't assume $gamma = delta = 1$
    $endgroup$
    – fleablood
    Jan 10 at 6:24
















4












$begingroup$


Given that $a$ and $b$ are real constants and that the equation $x^4+ax^3+2x^2+bx+1=0$ has at least one real root, find the minimum possible value of $a^2+b^2$.



I began this way: Let the polynomial be factorized as $(x^2+alpha x + 1)(x^2+beta x +1)$. Then expanding and comparing coefficients we get $alphabeta=0$, meaning either $alpha=0$ or $beta=0$. Suppose $alpha=0$. Then we see that $(x^2+beta x+1)$ should have real roots, from which we get $beta^2 geq 4$. But we get $a=b=beta$ from the comparison above. So $a^2+b^2 = 2beta^2 geq 8$.



Is it correct? Or is there any mistake? Any other solution is also welcome.






polynomials







share|cite|improve this question









$endgroup$








  • 2




    $begingroup$
    How do you know the polynomial has such a factorisation?
    $endgroup$
    – Lord Shark the Unknown
    Jan 10 at 5:27










  • $begingroup$
    Any fourth degree polynomial can be factorised into product of two quadratic expressions, no matter what the nature of its roots are, right?
    $endgroup$
    – Yellow
    Jan 10 at 5:32










  • $begingroup$
    @Any Radha, yes but not in that way you write...
    $endgroup$
    – dmtri
    Jan 10 at 5:38










  • $begingroup$
    I can't find anything wrong in your solution.
    $endgroup$
    – Anik Bhowmick
    Jan 10 at 5:43






  • 1




    $begingroup$
    Wouldn't this imply that $alpha + beta = a = b$ which need not be the case? If $x^4 + ax^3 + 2x^2 +bx +1 = (m x^2 + nx + p)(jx^2 + kx + q)$ we know $j = frac 1m$ and wolog it factors to $(x^2 + alpha x + gamma)(x^2 + beta x + delta)$ but we can't assume $gamma = delta = 1$
    $endgroup$
    – fleablood
    Jan 10 at 6:24














4












4








4


2



$begingroup$


Given that $a$ and $b$ are real constants and that the equation $x^4+ax^3+2x^2+bx+1=0$ has at least one real root, find the minimum possible value of $a^2+b^2$.



I began this way: Let the polynomial be factorized as $(x^2+alpha x + 1)(x^2+beta x +1)$. Then expanding and comparing coefficients we get $alphabeta=0$, meaning either $alpha=0$ or $beta=0$. Suppose $alpha=0$. Then we see that $(x^2+beta x+1)$ should have real roots, from which we get $beta^2 geq 4$. But we get $a=b=beta$ from the comparison above. So $a^2+b^2 = 2beta^2 geq 8$.



Is it correct? Or is there any mistake? Any other solution is also welcome.






polynomials







share|cite|improve this question









$endgroup$




Given that $a$ and $b$ are real constants and that the equation $x^4+ax^3+2x^2+bx+1=0$ has at least one real root, find the minimum possible value of $a^2+b^2$.



I began this way: Let the polynomial be factorized as $(x^2+alpha x + 1)(x^2+beta x +1)$. Then expanding and comparing coefficients we get $alphabeta=0$, meaning either $alpha=0$ or $beta=0$. Suppose $alpha=0$. Then we see that $(x^2+beta x+1)$ should have real roots, from which we get $beta^2 geq 4$. But we get $a=b=beta$ from the comparison above. So $a^2+b^2 = 2beta^2 geq 8$.



Is it correct? Or is there any mistake? Any other solution is also welcome.






polynomials




polynomials






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 10 at 5:24









YellowYellow

16011




16011








  • 2




    $begingroup$
    How do you know the polynomial has such a factorisation?
    $endgroup$
    – Lord Shark the Unknown
    Jan 10 at 5:27










  • $begingroup$
    Any fourth degree polynomial can be factorised into product of two quadratic expressions, no matter what the nature of its roots are, right?
    $endgroup$
    – Yellow
    Jan 10 at 5:32










  • $begingroup$
    @Any Radha, yes but not in that way you write...
    $endgroup$
    – dmtri
    Jan 10 at 5:38










  • $begingroup$
    I can't find anything wrong in your solution.
    $endgroup$
    – Anik Bhowmick
    Jan 10 at 5:43






  • 1




    $begingroup$
    Wouldn't this imply that $alpha + beta = a = b$ which need not be the case? If $x^4 + ax^3 + 2x^2 +bx +1 = (m x^2 + nx + p)(jx^2 + kx + q)$ we know $j = frac 1m$ and wolog it factors to $(x^2 + alpha x + gamma)(x^2 + beta x + delta)$ but we can't assume $gamma = delta = 1$
    $endgroup$
    – fleablood
    Jan 10 at 6:24














  • 2




    $begingroup$
    How do you know the polynomial has such a factorisation?
    $endgroup$
    – Lord Shark the Unknown
    Jan 10 at 5:27










  • $begingroup$
    Any fourth degree polynomial can be factorised into product of two quadratic expressions, no matter what the nature of its roots are, right?
    $endgroup$
    – Yellow
    Jan 10 at 5:32










  • $begingroup$
    @Any Radha, yes but not in that way you write...
    $endgroup$
    – dmtri
    Jan 10 at 5:38










  • $begingroup$
    I can't find anything wrong in your solution.
    $endgroup$
    – Anik Bhowmick
    Jan 10 at 5:43






  • 1




    $begingroup$
    Wouldn't this imply that $alpha + beta = a = b$ which need not be the case? If $x^4 + ax^3 + 2x^2 +bx +1 = (m x^2 + nx + p)(jx^2 + kx + q)$ we know $j = frac 1m$ and wolog it factors to $(x^2 + alpha x + gamma)(x^2 + beta x + delta)$ but we can't assume $gamma = delta = 1$
    $endgroup$
    – fleablood
    Jan 10 at 6:24








2




2




$begingroup$
How do you know the polynomial has such a factorisation?
$endgroup$
– Lord Shark the Unknown
Jan 10 at 5:27




$begingroup$
How do you know the polynomial has such a factorisation?
$endgroup$
– Lord Shark the Unknown
Jan 10 at 5:27












$begingroup$
Any fourth degree polynomial can be factorised into product of two quadratic expressions, no matter what the nature of its roots are, right?
$endgroup$
– Yellow
Jan 10 at 5:32




$begingroup$
Any fourth degree polynomial can be factorised into product of two quadratic expressions, no matter what the nature of its roots are, right?
$endgroup$
– Yellow
Jan 10 at 5:32












$begingroup$
@Any Radha, yes but not in that way you write...
$endgroup$
– dmtri
Jan 10 at 5:38




$begingroup$
@Any Radha, yes but not in that way you write...
$endgroup$
– dmtri
Jan 10 at 5:38












$begingroup$
I can't find anything wrong in your solution.
$endgroup$
– Anik Bhowmick
Jan 10 at 5:43




$begingroup$
I can't find anything wrong in your solution.
$endgroup$
– Anik Bhowmick
Jan 10 at 5:43




1




1




$begingroup$
Wouldn't this imply that $alpha + beta = a = b$ which need not be the case? If $x^4 + ax^3 + 2x^2 +bx +1 = (m x^2 + nx + p)(jx^2 + kx + q)$ we know $j = frac 1m$ and wolog it factors to $(x^2 + alpha x + gamma)(x^2 + beta x + delta)$ but we can't assume $gamma = delta = 1$
$endgroup$
– fleablood
Jan 10 at 6:24




$begingroup$
Wouldn't this imply that $alpha + beta = a = b$ which need not be the case? If $x^4 + ax^3 + 2x^2 +bx +1 = (m x^2 + nx + p)(jx^2 + kx + q)$ we know $j = frac 1m$ and wolog it factors to $(x^2 + alpha x + gamma)(x^2 + beta x + delta)$ but we can't assume $gamma = delta = 1$
$endgroup$
– fleablood
Jan 10 at 6:24










2 Answers
2






active

oldest

votes


















6












$begingroup$

$$x^4+ax^3+2x^2+bx+1=$$
$$=(x^2+frac{a}{2}x)^2-frac{a^2}{4}x^2+2x^2+bx+1=$$
$$=(x^2+frac{a}{2}x)^2-frac{a^2}{4}x^2+2x^2-frac{b^2}{4}x^2+(frac{b}{2}x+1)^2=$$
$$=(x^2+frac{a}{2}x)^2+x^2(2-frac{a^2+b^2}{4})+(frac{b}{2}x+1)^2$$



If $2-frac{a^2+b^2}{4}>0$, then we have a problem; the sum of three nonnegative terms equals zero (at our real root), which can only happen if $x=0$ to make the second term zero. But then $(frac{b}{2}x+1)^2>0$. So if $a^2+b^2<8$ there is no real root. Hence $a^2+b^2ge 8$.



Now, if $a^2+b^2=8$, then the middle term is zero. Can we make the other two terms zero as well? We need $x=-frac{a}{2}$ and $x=-frac{2}{b}$, to make the first and last terms zero. This can be achieved, with $a=b=2$ and $x=-1$.



Hence, the minimum possible value of $a^2+b^2$ is indeed $8$.






share|cite|improve this answer









$endgroup$









  • 2




    $begingroup$
    Whoa, that was quick..
    $endgroup$
    – Yellow
    Jan 10 at 5:42






  • 2




    $begingroup$
    @AnuRadha, I've seen similar problems before, involving completing the square repeatedly.
    $endgroup$
    – vadim123
    Jan 10 at 5:49



















1












$begingroup$

I would write first $(x^2+cx+d) (x^2+ex+f) $ and then find $c, d, e, f$ in terms of $a, b$.






share|cite|improve this answer









$endgroup$













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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    6












    $begingroup$

    $$x^4+ax^3+2x^2+bx+1=$$
    $$=(x^2+frac{a}{2}x)^2-frac{a^2}{4}x^2+2x^2+bx+1=$$
    $$=(x^2+frac{a}{2}x)^2-frac{a^2}{4}x^2+2x^2-frac{b^2}{4}x^2+(frac{b}{2}x+1)^2=$$
    $$=(x^2+frac{a}{2}x)^2+x^2(2-frac{a^2+b^2}{4})+(frac{b}{2}x+1)^2$$



    If $2-frac{a^2+b^2}{4}>0$, then we have a problem; the sum of three nonnegative terms equals zero (at our real root), which can only happen if $x=0$ to make the second term zero. But then $(frac{b}{2}x+1)^2>0$. So if $a^2+b^2<8$ there is no real root. Hence $a^2+b^2ge 8$.



    Now, if $a^2+b^2=8$, then the middle term is zero. Can we make the other two terms zero as well? We need $x=-frac{a}{2}$ and $x=-frac{2}{b}$, to make the first and last terms zero. This can be achieved, with $a=b=2$ and $x=-1$.



    Hence, the minimum possible value of $a^2+b^2$ is indeed $8$.






    share|cite|improve this answer









    $endgroup$









    • 2




      $begingroup$
      Whoa, that was quick..
      $endgroup$
      – Yellow
      Jan 10 at 5:42






    • 2




      $begingroup$
      @AnuRadha, I've seen similar problems before, involving completing the square repeatedly.
      $endgroup$
      – vadim123
      Jan 10 at 5:49
















    6












    $begingroup$

    $$x^4+ax^3+2x^2+bx+1=$$
    $$=(x^2+frac{a}{2}x)^2-frac{a^2}{4}x^2+2x^2+bx+1=$$
    $$=(x^2+frac{a}{2}x)^2-frac{a^2}{4}x^2+2x^2-frac{b^2}{4}x^2+(frac{b}{2}x+1)^2=$$
    $$=(x^2+frac{a}{2}x)^2+x^2(2-frac{a^2+b^2}{4})+(frac{b}{2}x+1)^2$$



    If $2-frac{a^2+b^2}{4}>0$, then we have a problem; the sum of three nonnegative terms equals zero (at our real root), which can only happen if $x=0$ to make the second term zero. But then $(frac{b}{2}x+1)^2>0$. So if $a^2+b^2<8$ there is no real root. Hence $a^2+b^2ge 8$.



    Now, if $a^2+b^2=8$, then the middle term is zero. Can we make the other two terms zero as well? We need $x=-frac{a}{2}$ and $x=-frac{2}{b}$, to make the first and last terms zero. This can be achieved, with $a=b=2$ and $x=-1$.



    Hence, the minimum possible value of $a^2+b^2$ is indeed $8$.






    share|cite|improve this answer









    $endgroup$









    • 2




      $begingroup$
      Whoa, that was quick..
      $endgroup$
      – Yellow
      Jan 10 at 5:42






    • 2




      $begingroup$
      @AnuRadha, I've seen similar problems before, involving completing the square repeatedly.
      $endgroup$
      – vadim123
      Jan 10 at 5:49














    6












    6








    6





    $begingroup$

    $$x^4+ax^3+2x^2+bx+1=$$
    $$=(x^2+frac{a}{2}x)^2-frac{a^2}{4}x^2+2x^2+bx+1=$$
    $$=(x^2+frac{a}{2}x)^2-frac{a^2}{4}x^2+2x^2-frac{b^2}{4}x^2+(frac{b}{2}x+1)^2=$$
    $$=(x^2+frac{a}{2}x)^2+x^2(2-frac{a^2+b^2}{4})+(frac{b}{2}x+1)^2$$



    If $2-frac{a^2+b^2}{4}>0$, then we have a problem; the sum of three nonnegative terms equals zero (at our real root), which can only happen if $x=0$ to make the second term zero. But then $(frac{b}{2}x+1)^2>0$. So if $a^2+b^2<8$ there is no real root. Hence $a^2+b^2ge 8$.



    Now, if $a^2+b^2=8$, then the middle term is zero. Can we make the other two terms zero as well? We need $x=-frac{a}{2}$ and $x=-frac{2}{b}$, to make the first and last terms zero. This can be achieved, with $a=b=2$ and $x=-1$.



    Hence, the minimum possible value of $a^2+b^2$ is indeed $8$.






    share|cite|improve this answer









    $endgroup$



    $$x^4+ax^3+2x^2+bx+1=$$
    $$=(x^2+frac{a}{2}x)^2-frac{a^2}{4}x^2+2x^2+bx+1=$$
    $$=(x^2+frac{a}{2}x)^2-frac{a^2}{4}x^2+2x^2-frac{b^2}{4}x^2+(frac{b}{2}x+1)^2=$$
    $$=(x^2+frac{a}{2}x)^2+x^2(2-frac{a^2+b^2}{4})+(frac{b}{2}x+1)^2$$



    If $2-frac{a^2+b^2}{4}>0$, then we have a problem; the sum of three nonnegative terms equals zero (at our real root), which can only happen if $x=0$ to make the second term zero. But then $(frac{b}{2}x+1)^2>0$. So if $a^2+b^2<8$ there is no real root. Hence $a^2+b^2ge 8$.



    Now, if $a^2+b^2=8$, then the middle term is zero. Can we make the other two terms zero as well? We need $x=-frac{a}{2}$ and $x=-frac{2}{b}$, to make the first and last terms zero. This can be achieved, with $a=b=2$ and $x=-1$.



    Hence, the minimum possible value of $a^2+b^2$ is indeed $8$.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Jan 10 at 5:38









    vadim123vadim123

    75.9k897189




    75.9k897189








    • 2




      $begingroup$
      Whoa, that was quick..
      $endgroup$
      – Yellow
      Jan 10 at 5:42






    • 2




      $begingroup$
      @AnuRadha, I've seen similar problems before, involving completing the square repeatedly.
      $endgroup$
      – vadim123
      Jan 10 at 5:49














    • 2




      $begingroup$
      Whoa, that was quick..
      $endgroup$
      – Yellow
      Jan 10 at 5:42






    • 2




      $begingroup$
      @AnuRadha, I've seen similar problems before, involving completing the square repeatedly.
      $endgroup$
      – vadim123
      Jan 10 at 5:49








    2




    2




    $begingroup$
    Whoa, that was quick..
    $endgroup$
    – Yellow
    Jan 10 at 5:42




    $begingroup$
    Whoa, that was quick..
    $endgroup$
    – Yellow
    Jan 10 at 5:42




    2




    2




    $begingroup$
    @AnuRadha, I've seen similar problems before, involving completing the square repeatedly.
    $endgroup$
    – vadim123
    Jan 10 at 5:49




    $begingroup$
    @AnuRadha, I've seen similar problems before, involving completing the square repeatedly.
    $endgroup$
    – vadim123
    Jan 10 at 5:49











    1












    $begingroup$

    I would write first $(x^2+cx+d) (x^2+ex+f) $ and then find $c, d, e, f$ in terms of $a, b$.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      I would write first $(x^2+cx+d) (x^2+ex+f) $ and then find $c, d, e, f$ in terms of $a, b$.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        I would write first $(x^2+cx+d) (x^2+ex+f) $ and then find $c, d, e, f$ in terms of $a, b$.






        share|cite|improve this answer









        $endgroup$



        I would write first $(x^2+cx+d) (x^2+ex+f) $ and then find $c, d, e, f$ in terms of $a, b$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 10 at 5:53









        dmtridmtri

        1,4542521




        1,4542521






























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            What does “Dominus providebit” mean?

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            5 Does "Dominus providebit" mean "The Lord will provide" or "The Lord provides"? I once had a bit of an argument with a guy who studied Latin over that. It's an inscription that appears on the rim of some old Swiss coins that I have. meaning latin-to-english-translation grammar-identification share | improve this question edited Jan 18 at 15:27 V2Blast 105 4 asked Jan 17 at 22:49 Donald K. Donald K. 26 1 ...
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