If $a$ and $b$ are integers s.t. $2x^2 -ax + 2 > 0$ and $x^2 -b x + 8 geq 0$ for all real numbers $x$,...












0












$begingroup$



If a and b are integers such that $2x^2-ax+2>0$ and $x^2-bx+8≥0$ for all real numbers $x$, then the largest possible value of $2a−6b$ is




Answer: 36.



My attempt: Multiply the first inequality by 2 and second by 6. This gave $4x^2-2ax+2>0$ and $6x^2-6bx+48≥0$. Subtracting the second inequality from the first one, we get $-2ax+6bx≥2x^2+44$. This means $2a-6b≤(2x^2+44)/-x.$ Now how to move further. Am I going in the right direction?






inequality polynomials quadratics







share|cite|improve this question











$endgroup$












  • $begingroup$
    You can't subtract inequalities at all...
    $endgroup$
    – Lucas Henrique
    Dec 19 '18 at 18:19










  • $begingroup$
    Thanks for this. Could you explain as to why can't we subtract inequalities?
    $endgroup$
    – Sherlock Watson
    Dec 19 '18 at 18:29










  • $begingroup$
    $2>1 land 8>7 not implies -6 = 2-8 > 1-7 = -6$.
    $endgroup$
    – Lucas Henrique
    Dec 19 '18 at 18:38










  • $begingroup$
    Ok. Got the idea. Thanks.
    $endgroup$
    – Sherlock Watson
    Dec 19 '18 at 18:44










  • $begingroup$
    Actually you can subtract inequalities but you must subtract the opposite side. If $a < b$ and $c < d$ then $a - d < b - c$. You can do that because taking away a LARGER thing gives a smaller result.
    $endgroup$
    – fleablood
    Dec 20 '18 at 23:35
















0












$begingroup$



If a and b are integers such that $2x^2-ax+2>0$ and $x^2-bx+8≥0$ for all real numbers $x$, then the largest possible value of $2a−6b$ is




Answer: 36.



My attempt: Multiply the first inequality by 2 and second by 6. This gave $4x^2-2ax+2>0$ and $6x^2-6bx+48≥0$. Subtracting the second inequality from the first one, we get $-2ax+6bx≥2x^2+44$. This means $2a-6b≤(2x^2+44)/-x.$ Now how to move further. Am I going in the right direction?






inequality polynomials quadratics







share|cite|improve this question











$endgroup$












  • $begingroup$
    You can't subtract inequalities at all...
    $endgroup$
    – Lucas Henrique
    Dec 19 '18 at 18:19










  • $begingroup$
    Thanks for this. Could you explain as to why can't we subtract inequalities?
    $endgroup$
    – Sherlock Watson
    Dec 19 '18 at 18:29










  • $begingroup$
    $2>1 land 8>7 not implies -6 = 2-8 > 1-7 = -6$.
    $endgroup$
    – Lucas Henrique
    Dec 19 '18 at 18:38










  • $begingroup$
    Ok. Got the idea. Thanks.
    $endgroup$
    – Sherlock Watson
    Dec 19 '18 at 18:44










  • $begingroup$
    Actually you can subtract inequalities but you must subtract the opposite side. If $a < b$ and $c < d$ then $a - d < b - c$. You can do that because taking away a LARGER thing gives a smaller result.
    $endgroup$
    – fleablood
    Dec 20 '18 at 23:35














0












0








0





$begingroup$



If a and b are integers such that $2x^2-ax+2>0$ and $x^2-bx+8≥0$ for all real numbers $x$, then the largest possible value of $2a−6b$ is




Answer: 36.



My attempt: Multiply the first inequality by 2 and second by 6. This gave $4x^2-2ax+2>0$ and $6x^2-6bx+48≥0$. Subtracting the second inequality from the first one, we get $-2ax+6bx≥2x^2+44$. This means $2a-6b≤(2x^2+44)/-x.$ Now how to move further. Am I going in the right direction?






inequality polynomials quadratics







share|cite|improve this question











$endgroup$





If a and b are integers such that $2x^2-ax+2>0$ and $x^2-bx+8≥0$ for all real numbers $x$, then the largest possible value of $2a−6b$ is




Answer: 36.



My attempt: Multiply the first inequality by 2 and second by 6. This gave $4x^2-2ax+2>0$ and $6x^2-6bx+48≥0$. Subtracting the second inequality from the first one, we get $-2ax+6bx≥2x^2+44$. This means $2a-6b≤(2x^2+44)/-x.$ Now how to move further. Am I going in the right direction?






inequality polynomials quadratics




inequality polynomials quadratics






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 19 '18 at 18:35









Maria Mazur

49.6k1361124




49.6k1361124










asked Dec 19 '18 at 18:17









Sherlock WatsonSherlock Watson

3462413




3462413












  • $begingroup$
    You can't subtract inequalities at all...
    $endgroup$
    – Lucas Henrique
    Dec 19 '18 at 18:19










  • $begingroup$
    Thanks for this. Could you explain as to why can't we subtract inequalities?
    $endgroup$
    – Sherlock Watson
    Dec 19 '18 at 18:29










  • $begingroup$
    $2>1 land 8>7 not implies -6 = 2-8 > 1-7 = -6$.
    $endgroup$
    – Lucas Henrique
    Dec 19 '18 at 18:38










  • $begingroup$
    Ok. Got the idea. Thanks.
    $endgroup$
    – Sherlock Watson
    Dec 19 '18 at 18:44










  • $begingroup$
    Actually you can subtract inequalities but you must subtract the opposite side. If $a < b$ and $c < d$ then $a - d < b - c$. You can do that because taking away a LARGER thing gives a smaller result.
    $endgroup$
    – fleablood
    Dec 20 '18 at 23:35


















  • $begingroup$
    You can't subtract inequalities at all...
    $endgroup$
    – Lucas Henrique
    Dec 19 '18 at 18:19










  • $begingroup$
    Thanks for this. Could you explain as to why can't we subtract inequalities?
    $endgroup$
    – Sherlock Watson
    Dec 19 '18 at 18:29










  • $begingroup$
    $2>1 land 8>7 not implies -6 = 2-8 > 1-7 = -6$.
    $endgroup$
    – Lucas Henrique
    Dec 19 '18 at 18:38










  • $begingroup$
    Ok. Got the idea. Thanks.
    $endgroup$
    – Sherlock Watson
    Dec 19 '18 at 18:44










  • $begingroup$
    Actually you can subtract inequalities but you must subtract the opposite side. If $a < b$ and $c < d$ then $a - d < b - c$. You can do that because taking away a LARGER thing gives a smaller result.
    $endgroup$
    – fleablood
    Dec 20 '18 at 23:35
















$begingroup$
You can't subtract inequalities at all...
$endgroup$
– Lucas Henrique
Dec 19 '18 at 18:19




$begingroup$
You can't subtract inequalities at all...
$endgroup$
– Lucas Henrique
Dec 19 '18 at 18:19












$begingroup$
Thanks for this. Could you explain as to why can't we subtract inequalities?
$endgroup$
– Sherlock Watson
Dec 19 '18 at 18:29




$begingroup$
Thanks for this. Could you explain as to why can't we subtract inequalities?
$endgroup$
– Sherlock Watson
Dec 19 '18 at 18:29












$begingroup$
$2>1 land 8>7 not implies -6 = 2-8 > 1-7 = -6$.
$endgroup$
– Lucas Henrique
Dec 19 '18 at 18:38




$begingroup$
$2>1 land 8>7 not implies -6 = 2-8 > 1-7 = -6$.
$endgroup$
– Lucas Henrique
Dec 19 '18 at 18:38












$begingroup$
Ok. Got the idea. Thanks.
$endgroup$
– Sherlock Watson
Dec 19 '18 at 18:44




$begingroup$
Ok. Got the idea. Thanks.
$endgroup$
– Sherlock Watson
Dec 19 '18 at 18:44












$begingroup$
Actually you can subtract inequalities but you must subtract the opposite side. If $a < b$ and $c < d$ then $a - d < b - c$. You can do that because taking away a LARGER thing gives a smaller result.
$endgroup$
– fleablood
Dec 20 '18 at 23:35




$begingroup$
Actually you can subtract inequalities but you must subtract the opposite side. If $a < b$ and $c < d$ then $a - d < b - c$. You can do that because taking away a LARGER thing gives a smaller result.
$endgroup$
– fleablood
Dec 20 '18 at 23:35










3 Answers
3






active

oldest

votes


















2












$begingroup$

The discriminant for first must be $<0$ and for second $leq 0$ so $$a^2-16<0;;;;{rm and};;;;b^2-32leq 0$$



So $$ aleq 3;;;;{rm and};;;;bgeq -5 implies 2a-6bleq 36$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    How do you determine that the discriminant has to be <0 and ≤0 for the first and second respectively?
    $endgroup$
    – Sherlock Watson
    Dec 19 '18 at 18:31






  • 1




    $begingroup$
    Because of > in first and $geq$ second inequality
    $endgroup$
    – Maria Mazur
    Dec 19 '18 at 18:33






  • 2




    $begingroup$
    That is in first case your parabole can not go below x-axsis, so it has no zeroes, that is if disciriminat is <.
    $endgroup$
    – Maria Mazur
    Dec 19 '18 at 18:34






  • 1




    $begingroup$
    In second case parabola can touch x-axsis so it can have at most one root
    $endgroup$
    – Maria Mazur
    Dec 19 '18 at 18:34










  • $begingroup$
    I understood this from this. quora.com/…
    $endgroup$
    – Sherlock Watson
    Dec 20 '18 at 2:23



















2












$begingroup$

Since these quadratic functions have a positive leading coefficient, they have a minimum on $bf R$, and we know the extremum of a quadratic function $p(x)=Ax^2+Bx+C$ is attained at $x=-dfrac B{2A}$, so the function is positive (resp. non-negative) for all $x$ if and only if $A>0$ and $pBigl(-dfrac B{2A}Bigr)>0:$ (resp. $ge 0$).



Here the hypotheses yield
begin{cases}2Bigl(dfrac a4Bigr)^2-dfrac{a^2}4+2>iff2-dfrac{a^2}8>0iff a^2<16, \[1ex]
Bigl(dfrac b2Bigr)^2-dfrac{b^2}2+8ge 0iff 8-dfrac{b^2}4ge 0iff b^2le 32,
end{cases}
Therefore we have $-4<a<4;$ and $;-4sqrt 2 le ble 4sqrt 2$, so that
$$left.begin{matrix}-8<2a<8\-24sqrt 2le -6ble24sqrt2end{matrix}right}Rightarrow -8(1+3sqrt2)<2a-6b<8(1+3sqrt 2).$$



Added:



This solution is for real $a$ and $b$. If ˆ$a$ and $b$ are constrained to be integers, we have the same basic inequalities with $a^2$ and $b^2$, but now we have the equivalence
$$begin{cases}
a^2<16\b^2le 32end{cases}iff begin{cases}
|a|<4\ |b|le 5end{cases}iff begin{cases}
-3le a le 3\ -5le ble 5end{cases}iff begin{cases}
-6le 2a le 6\ -30le -6ble 30end{cases}$$
so $;2a-6ble 36$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    That would be a good approach if $a,b$ being integers wasn’t a condition. However in this case... stepping down to 36 from 41.94... won’t be justified.
    $endgroup$
    – Macavity
    Dec 20 '18 at 18:58










  • $begingroup$
    @Macavity: I've added the very simple conclusion for integers.
    $endgroup$
    – Bernard
    Dec 20 '18 at 19:20










  • $begingroup$
    Yes, of course that conclusion works.. +1
    $endgroup$
    – Macavity
    Dec 20 '18 at 19:26



















1












$begingroup$

As pointed out in the comments, you cannot subtract inequalities since $a>0$ & $b>0$ does not imply $a-b>0$ (eg. $7>0$ & $8>0$ but $7-8<0$). You should go forward with such problems involving quadratic equations using discriminants as used in the solution above by @greedoid.






share|cite|improve this answer









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    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2












    $begingroup$

    The discriminant for first must be $<0$ and for second $leq 0$ so $$a^2-16<0;;;;{rm and};;;;b^2-32leq 0$$



    So $$ aleq 3;;;;{rm and};;;;bgeq -5 implies 2a-6bleq 36$$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      How do you determine that the discriminant has to be <0 and ≤0 for the first and second respectively?
      $endgroup$
      – Sherlock Watson
      Dec 19 '18 at 18:31






    • 1




      $begingroup$
      Because of > in first and $geq$ second inequality
      $endgroup$
      – Maria Mazur
      Dec 19 '18 at 18:33






    • 2




      $begingroup$
      That is in first case your parabole can not go below x-axsis, so it has no zeroes, that is if disciriminat is <.
      $endgroup$
      – Maria Mazur
      Dec 19 '18 at 18:34






    • 1




      $begingroup$
      In second case parabola can touch x-axsis so it can have at most one root
      $endgroup$
      – Maria Mazur
      Dec 19 '18 at 18:34










    • $begingroup$
      I understood this from this. quora.com/…
      $endgroup$
      – Sherlock Watson
      Dec 20 '18 at 2:23
















    2












    $begingroup$

    The discriminant for first must be $<0$ and for second $leq 0$ so $$a^2-16<0;;;;{rm and};;;;b^2-32leq 0$$



    So $$ aleq 3;;;;{rm and};;;;bgeq -5 implies 2a-6bleq 36$$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      How do you determine that the discriminant has to be <0 and ≤0 for the first and second respectively?
      $endgroup$
      – Sherlock Watson
      Dec 19 '18 at 18:31






    • 1




      $begingroup$
      Because of > in first and $geq$ second inequality
      $endgroup$
      – Maria Mazur
      Dec 19 '18 at 18:33






    • 2




      $begingroup$
      That is in first case your parabole can not go below x-axsis, so it has no zeroes, that is if disciriminat is <.
      $endgroup$
      – Maria Mazur
      Dec 19 '18 at 18:34






    • 1




      $begingroup$
      In second case parabola can touch x-axsis so it can have at most one root
      $endgroup$
      – Maria Mazur
      Dec 19 '18 at 18:34










    • $begingroup$
      I understood this from this. quora.com/…
      $endgroup$
      – Sherlock Watson
      Dec 20 '18 at 2:23














    2












    2








    2





    $begingroup$

    The discriminant for first must be $<0$ and for second $leq 0$ so $$a^2-16<0;;;;{rm and};;;;b^2-32leq 0$$



    So $$ aleq 3;;;;{rm and};;;;bgeq -5 implies 2a-6bleq 36$$






    share|cite|improve this answer











    $endgroup$



    The discriminant for first must be $<0$ and for second $leq 0$ so $$a^2-16<0;;;;{rm and};;;;b^2-32leq 0$$



    So $$ aleq 3;;;;{rm and};;;;bgeq -5 implies 2a-6bleq 36$$







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Dec 19 '18 at 18:56









    Bernard

    123k741117




    123k741117










    answered Dec 19 '18 at 18:22









    Maria MazurMaria Mazur

    49.6k1361124




    49.6k1361124












    • $begingroup$
      How do you determine that the discriminant has to be <0 and ≤0 for the first and second respectively?
      $endgroup$
      – Sherlock Watson
      Dec 19 '18 at 18:31






    • 1




      $begingroup$
      Because of > in first and $geq$ second inequality
      $endgroup$
      – Maria Mazur
      Dec 19 '18 at 18:33






    • 2




      $begingroup$
      That is in first case your parabole can not go below x-axsis, so it has no zeroes, that is if disciriminat is <.
      $endgroup$
      – Maria Mazur
      Dec 19 '18 at 18:34






    • 1




      $begingroup$
      In second case parabola can touch x-axsis so it can have at most one root
      $endgroup$
      – Maria Mazur
      Dec 19 '18 at 18:34










    • $begingroup$
      I understood this from this. quora.com/…
      $endgroup$
      – Sherlock Watson
      Dec 20 '18 at 2:23


















    • $begingroup$
      How do you determine that the discriminant has to be <0 and ≤0 for the first and second respectively?
      $endgroup$
      – Sherlock Watson
      Dec 19 '18 at 18:31






    • 1




      $begingroup$
      Because of > in first and $geq$ second inequality
      $endgroup$
      – Maria Mazur
      Dec 19 '18 at 18:33






    • 2




      $begingroup$
      That is in first case your parabole can not go below x-axsis, so it has no zeroes, that is if disciriminat is <.
      $endgroup$
      – Maria Mazur
      Dec 19 '18 at 18:34






    • 1




      $begingroup$
      In second case parabola can touch x-axsis so it can have at most one root
      $endgroup$
      – Maria Mazur
      Dec 19 '18 at 18:34










    • $begingroup$
      I understood this from this. quora.com/…
      $endgroup$
      – Sherlock Watson
      Dec 20 '18 at 2:23
















    $begingroup$
    How do you determine that the discriminant has to be <0 and ≤0 for the first and second respectively?
    $endgroup$
    – Sherlock Watson
    Dec 19 '18 at 18:31




    $begingroup$
    How do you determine that the discriminant has to be <0 and ≤0 for the first and second respectively?
    $endgroup$
    – Sherlock Watson
    Dec 19 '18 at 18:31




    1




    1




    $begingroup$
    Because of > in first and $geq$ second inequality
    $endgroup$
    – Maria Mazur
    Dec 19 '18 at 18:33




    $begingroup$
    Because of > in first and $geq$ second inequality
    $endgroup$
    – Maria Mazur
    Dec 19 '18 at 18:33




    2




    2




    $begingroup$
    That is in first case your parabole can not go below x-axsis, so it has no zeroes, that is if disciriminat is <.
    $endgroup$
    – Maria Mazur
    Dec 19 '18 at 18:34




    $begingroup$
    That is in first case your parabole can not go below x-axsis, so it has no zeroes, that is if disciriminat is <.
    $endgroup$
    – Maria Mazur
    Dec 19 '18 at 18:34




    1




    1




    $begingroup$
    In second case parabola can touch x-axsis so it can have at most one root
    $endgroup$
    – Maria Mazur
    Dec 19 '18 at 18:34




    $begingroup$
    In second case parabola can touch x-axsis so it can have at most one root
    $endgroup$
    – Maria Mazur
    Dec 19 '18 at 18:34












    $begingroup$
    I understood this from this. quora.com/…
    $endgroup$
    – Sherlock Watson
    Dec 20 '18 at 2:23




    $begingroup$
    I understood this from this. quora.com/…
    $endgroup$
    – Sherlock Watson
    Dec 20 '18 at 2:23











    2












    $begingroup$

    Since these quadratic functions have a positive leading coefficient, they have a minimum on $bf R$, and we know the extremum of a quadratic function $p(x)=Ax^2+Bx+C$ is attained at $x=-dfrac B{2A}$, so the function is positive (resp. non-negative) for all $x$ if and only if $A>0$ and $pBigl(-dfrac B{2A}Bigr)>0:$ (resp. $ge 0$).



    Here the hypotheses yield
    begin{cases}2Bigl(dfrac a4Bigr)^2-dfrac{a^2}4+2>iff2-dfrac{a^2}8>0iff a^2<16, \[1ex]
    Bigl(dfrac b2Bigr)^2-dfrac{b^2}2+8ge 0iff 8-dfrac{b^2}4ge 0iff b^2le 32,
    end{cases}
    Therefore we have $-4<a<4;$ and $;-4sqrt 2 le ble 4sqrt 2$, so that
    $$left.begin{matrix}-8<2a<8\-24sqrt 2le -6ble24sqrt2end{matrix}right}Rightarrow -8(1+3sqrt2)<2a-6b<8(1+3sqrt 2).$$



    Added:



    This solution is for real $a$ and $b$. If ˆ$a$ and $b$ are constrained to be integers, we have the same basic inequalities with $a^2$ and $b^2$, but now we have the equivalence
    $$begin{cases}
    a^2<16\b^2le 32end{cases}iff begin{cases}
    |a|<4\ |b|le 5end{cases}iff begin{cases}
    -3le a le 3\ -5le ble 5end{cases}iff begin{cases}
    -6le 2a le 6\ -30le -6ble 30end{cases}$$
    so $;2a-6ble 36$.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      That would be a good approach if $a,b$ being integers wasn’t a condition. However in this case... stepping down to 36 from 41.94... won’t be justified.
      $endgroup$
      – Macavity
      Dec 20 '18 at 18:58










    • $begingroup$
      @Macavity: I've added the very simple conclusion for integers.
      $endgroup$
      – Bernard
      Dec 20 '18 at 19:20










    • $begingroup$
      Yes, of course that conclusion works.. +1
      $endgroup$
      – Macavity
      Dec 20 '18 at 19:26
















    2












    $begingroup$

    Since these quadratic functions have a positive leading coefficient, they have a minimum on $bf R$, and we know the extremum of a quadratic function $p(x)=Ax^2+Bx+C$ is attained at $x=-dfrac B{2A}$, so the function is positive (resp. non-negative) for all $x$ if and only if $A>0$ and $pBigl(-dfrac B{2A}Bigr)>0:$ (resp. $ge 0$).



    Here the hypotheses yield
    begin{cases}2Bigl(dfrac a4Bigr)^2-dfrac{a^2}4+2>iff2-dfrac{a^2}8>0iff a^2<16, \[1ex]
    Bigl(dfrac b2Bigr)^2-dfrac{b^2}2+8ge 0iff 8-dfrac{b^2}4ge 0iff b^2le 32,
    end{cases}
    Therefore we have $-4<a<4;$ and $;-4sqrt 2 le ble 4sqrt 2$, so that
    $$left.begin{matrix}-8<2a<8\-24sqrt 2le -6ble24sqrt2end{matrix}right}Rightarrow -8(1+3sqrt2)<2a-6b<8(1+3sqrt 2).$$



    Added:



    This solution is for real $a$ and $b$. If ˆ$a$ and $b$ are constrained to be integers, we have the same basic inequalities with $a^2$ and $b^2$, but now we have the equivalence
    $$begin{cases}
    a^2<16\b^2le 32end{cases}iff begin{cases}
    |a|<4\ |b|le 5end{cases}iff begin{cases}
    -3le a le 3\ -5le ble 5end{cases}iff begin{cases}
    -6le 2a le 6\ -30le -6ble 30end{cases}$$
    so $;2a-6ble 36$.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      That would be a good approach if $a,b$ being integers wasn’t a condition. However in this case... stepping down to 36 from 41.94... won’t be justified.
      $endgroup$
      – Macavity
      Dec 20 '18 at 18:58










    • $begingroup$
      @Macavity: I've added the very simple conclusion for integers.
      $endgroup$
      – Bernard
      Dec 20 '18 at 19:20










    • $begingroup$
      Yes, of course that conclusion works.. +1
      $endgroup$
      – Macavity
      Dec 20 '18 at 19:26














    2












    2








    2





    $begingroup$

    Since these quadratic functions have a positive leading coefficient, they have a minimum on $bf R$, and we know the extremum of a quadratic function $p(x)=Ax^2+Bx+C$ is attained at $x=-dfrac B{2A}$, so the function is positive (resp. non-negative) for all $x$ if and only if $A>0$ and $pBigl(-dfrac B{2A}Bigr)>0:$ (resp. $ge 0$).



    Here the hypotheses yield
    begin{cases}2Bigl(dfrac a4Bigr)^2-dfrac{a^2}4+2>iff2-dfrac{a^2}8>0iff a^2<16, \[1ex]
    Bigl(dfrac b2Bigr)^2-dfrac{b^2}2+8ge 0iff 8-dfrac{b^2}4ge 0iff b^2le 32,
    end{cases}
    Therefore we have $-4<a<4;$ and $;-4sqrt 2 le ble 4sqrt 2$, so that
    $$left.begin{matrix}-8<2a<8\-24sqrt 2le -6ble24sqrt2end{matrix}right}Rightarrow -8(1+3sqrt2)<2a-6b<8(1+3sqrt 2).$$



    Added:



    This solution is for real $a$ and $b$. If ˆ$a$ and $b$ are constrained to be integers, we have the same basic inequalities with $a^2$ and $b^2$, but now we have the equivalence
    $$begin{cases}
    a^2<16\b^2le 32end{cases}iff begin{cases}
    |a|<4\ |b|le 5end{cases}iff begin{cases}
    -3le a le 3\ -5le ble 5end{cases}iff begin{cases}
    -6le 2a le 6\ -30le -6ble 30end{cases}$$
    so $;2a-6ble 36$.






    share|cite|improve this answer











    $endgroup$



    Since these quadratic functions have a positive leading coefficient, they have a minimum on $bf R$, and we know the extremum of a quadratic function $p(x)=Ax^2+Bx+C$ is attained at $x=-dfrac B{2A}$, so the function is positive (resp. non-negative) for all $x$ if and only if $A>0$ and $pBigl(-dfrac B{2A}Bigr)>0:$ (resp. $ge 0$).



    Here the hypotheses yield
    begin{cases}2Bigl(dfrac a4Bigr)^2-dfrac{a^2}4+2>iff2-dfrac{a^2}8>0iff a^2<16, \[1ex]
    Bigl(dfrac b2Bigr)^2-dfrac{b^2}2+8ge 0iff 8-dfrac{b^2}4ge 0iff b^2le 32,
    end{cases}
    Therefore we have $-4<a<4;$ and $;-4sqrt 2 le ble 4sqrt 2$, so that
    $$left.begin{matrix}-8<2a<8\-24sqrt 2le -6ble24sqrt2end{matrix}right}Rightarrow -8(1+3sqrt2)<2a-6b<8(1+3sqrt 2).$$



    Added:



    This solution is for real $a$ and $b$. If ˆ$a$ and $b$ are constrained to be integers, we have the same basic inequalities with $a^2$ and $b^2$, but now we have the equivalence
    $$begin{cases}
    a^2<16\b^2le 32end{cases}iff begin{cases}
    |a|<4\ |b|le 5end{cases}iff begin{cases}
    -3le a le 3\ -5le ble 5end{cases}iff begin{cases}
    -6le 2a le 6\ -30le -6ble 30end{cases}$$
    so $;2a-6ble 36$.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Dec 20 '18 at 19:19

























    answered Dec 19 '18 at 19:23









    BernardBernard

    123k741117




    123k741117












    • $begingroup$
      That would be a good approach if $a,b$ being integers wasn’t a condition. However in this case... stepping down to 36 from 41.94... won’t be justified.
      $endgroup$
      – Macavity
      Dec 20 '18 at 18:58










    • $begingroup$
      @Macavity: I've added the very simple conclusion for integers.
      $endgroup$
      – Bernard
      Dec 20 '18 at 19:20










    • $begingroup$
      Yes, of course that conclusion works.. +1
      $endgroup$
      – Macavity
      Dec 20 '18 at 19:26


















    • $begingroup$
      That would be a good approach if $a,b$ being integers wasn’t a condition. However in this case... stepping down to 36 from 41.94... won’t be justified.
      $endgroup$
      – Macavity
      Dec 20 '18 at 18:58










    • $begingroup$
      @Macavity: I've added the very simple conclusion for integers.
      $endgroup$
      – Bernard
      Dec 20 '18 at 19:20










    • $begingroup$
      Yes, of course that conclusion works.. +1
      $endgroup$
      – Macavity
      Dec 20 '18 at 19:26
















    $begingroup$
    That would be a good approach if $a,b$ being integers wasn’t a condition. However in this case... stepping down to 36 from 41.94... won’t be justified.
    $endgroup$
    – Macavity
    Dec 20 '18 at 18:58




    $begingroup$
    That would be a good approach if $a,b$ being integers wasn’t a condition. However in this case... stepping down to 36 from 41.94... won’t be justified.
    $endgroup$
    – Macavity
    Dec 20 '18 at 18:58












    $begingroup$
    @Macavity: I've added the very simple conclusion for integers.
    $endgroup$
    – Bernard
    Dec 20 '18 at 19:20




    $begingroup$
    @Macavity: I've added the very simple conclusion for integers.
    $endgroup$
    – Bernard
    Dec 20 '18 at 19:20












    $begingroup$
    Yes, of course that conclusion works.. +1
    $endgroup$
    – Macavity
    Dec 20 '18 at 19:26




    $begingroup$
    Yes, of course that conclusion works.. +1
    $endgroup$
    – Macavity
    Dec 20 '18 at 19:26











    1












    $begingroup$

    As pointed out in the comments, you cannot subtract inequalities since $a>0$ & $b>0$ does not imply $a-b>0$ (eg. $7>0$ & $8>0$ but $7-8<0$). You should go forward with such problems involving quadratic equations using discriminants as used in the solution above by @greedoid.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      As pointed out in the comments, you cannot subtract inequalities since $a>0$ & $b>0$ does not imply $a-b>0$ (eg. $7>0$ & $8>0$ but $7-8<0$). You should go forward with such problems involving quadratic equations using discriminants as used in the solution above by @greedoid.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        As pointed out in the comments, you cannot subtract inequalities since $a>0$ & $b>0$ does not imply $a-b>0$ (eg. $7>0$ & $8>0$ but $7-8<0$). You should go forward with such problems involving quadratic equations using discriminants as used in the solution above by @greedoid.






        share|cite|improve this answer









        $endgroup$



        As pointed out in the comments, you cannot subtract inequalities since $a>0$ & $b>0$ does not imply $a-b>0$ (eg. $7>0$ & $8>0$ but $7-8<0$). You should go forward with such problems involving quadratic equations using discriminants as used in the solution above by @greedoid.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 19 '18 at 18:35









        Kshitij MishraKshitij Mishra

        313




        313






























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