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 $x$,...
$begingroup$
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
edited Dec 19 '18 at 18:35
greedoid
42.7k1153105
asked Dec 19 '18 at 18:17
Sherlock WatsonSherlock Watson
3462413
$endgroup$
|
show 1 more comment
$begingroup$
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
edited Dec 19 '18 at 18:35
greedoid
42.7k1153105
asked Dec 19 '18 at 18:17
Sherlock WatsonSherlock Watson
3462413
$endgroup$
$begingroup$
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
|
show 1 more comment
$begingroup$
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
edited Dec 19 '18 at 18:35
greedoid
42.7k1153105
asked Dec 19 '18 at 18:17
Sherlock WatsonSherlock Watson
3462413
$endgroup$
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
edited Dec 19 '18 at 18:35
greedoid
42.7k1153105
asked Dec 19 '18 at 18:17
Sherlock WatsonSherlock Watson
3462413
edited Dec 19 '18 at 18:35
greedoid
42.7k1153105
asked Dec 19 '18 at 18:17
Sherlock WatsonSherlock Watson
3462413
edited Dec 19 '18 at 18:35
greedoid
42.7k1153105
edited Dec 19 '18 at 18:35
greedoid
42.7k1153105
edited Dec 19 '18 at 18:35
greedoid
42.7k1153105
42.7k1153105
asked Dec 19 '18 at 18:17
Sherlock WatsonSherlock Watson
3462413
asked Dec 19 '18 at 18:17
Sherlock WatsonSherlock Watson
3462413
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
|
show 1 more comment
$begingroup$
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
|
show 1 more comment
3 Answers
3
active
oldest
votes
$begingroup$
The discriminant for first must be $<0$ and for second $leq 0$ so $$a^2-16<0;;;;{rm and};;;;b^2-32leq 0$$
So $$ aleq 3;;;;{rm and};;;;bgeq -5 implies 2a-6bleq 36$$
edited Dec 19 '18 at 18:56
Bernard
121k740116
answered Dec 19 '18 at 18:22
greedoidgreedoid
42.7k1153105
$endgroup$
$begingroup$
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
$endgroup$
– greedoid
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 <.
$endgroup$
– greedoid
Dec 19 '18 at 18:34
1
$begingroup$
In second case parabola can touch x-axsis so it can have at most one root
$endgroup$
– greedoid
Dec 19 '18 at 18:34
$begingroup$
I understood this from this. quora.com/…
$endgroup$
– Sherlock Watson
Dec 20 '18 at 2:23
add a comment |
$begingroup$
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 $;2a-6ble 36$.
answered Dec 19 '18 at 19:23
BernardBernard
121k740116
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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.
answered Dec 19 '18 at 18:35
Kshitij MishraKshitij Mishra
413
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The discriminant for first must be $<0$ and for second $leq 0$ so $$a^2-16<0;;;;{rm and};;;;b^2-32leq 0$$
So $$ aleq 3;;;;{rm and};;;;bgeq -5 implies 2a-6bleq 36$$
edited Dec 19 '18 at 18:56
Bernard
121k740116
answered Dec 19 '18 at 18:22
greedoidgreedoid
42.7k1153105
$endgroup$
$begingroup$
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
$endgroup$
– greedoid
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 <.
$endgroup$
– greedoid
Dec 19 '18 at 18:34
1
$begingroup$
In second case parabola can touch x-axsis so it can have at most one root
$endgroup$
– greedoid
Dec 19 '18 at 18:34
$begingroup$
I understood this from this. quora.com/…
$endgroup$
– Sherlock Watson
Dec 20 '18 at 2:23
add a comment |
$begingroup$
The discriminant for first must be $<0$ and for second $leq 0$ so $$a^2-16<0;;;;{rm and};;;;b^2-32leq 0$$
So $$ aleq 3;;;;{rm and};;;;bgeq -5 implies 2a-6bleq 36$$
edited Dec 19 '18 at 18:56
Bernard
121k740116
answered Dec 19 '18 at 18:22
greedoidgreedoid
42.7k1153105
$endgroup$
$begingroup$
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
$endgroup$
– greedoid
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 <.
$endgroup$
– greedoid
Dec 19 '18 at 18:34
1
$begingroup$
In second case parabola can touch x-axsis so it can have at most one root
$endgroup$
– greedoid
Dec 19 '18 at 18:34
$begingroup$
I understood this from this. quora.com/…
$endgroup$
– Sherlock Watson
Dec 20 '18 at 2:23
add a comment |
$begingroup$
The discriminant for first must be $<0$ and for second $leq 0$ so $$a^2-16<0;;;;{rm and};;;;b^2-32leq 0$$
So $$ aleq 3;;;;{rm and};;;;bgeq -5 implies 2a-6bleq 36$$
edited Dec 19 '18 at 18:56
Bernard
121k740116
answered Dec 19 '18 at 18:22
greedoidgreedoid
42.7k1153105
$endgroup$
The discriminant for first must be $<0$ and for second $leq 0$ so $$a^2-16<0;;;;{rm and};;;;b^2-32leq 0$$
So $$ aleq 3;;;;{rm and};;;;bgeq -5 implies 2a-6bleq 36$$
edited Dec 19 '18 at 18:56
Bernard
121k740116
answered Dec 19 '18 at 18:22
greedoidgreedoid
42.7k1153105
edited Dec 19 '18 at 18:56
Bernard
121k740116
edited Dec 19 '18 at 18:56
Bernard
121k740116
edited Dec 19 '18 at 18:56
Bernard
121k740116
121k740116
answered Dec 19 '18 at 18:22
greedoidgreedoid
42.7k1153105
answered Dec 19 '18 at 18:22
greedoidgreedoid
42.7k1153105
answered Dec 19 '18 at 18:22
greedoidgreedoid
42.7k1153105
42.7k1153105
$begingroup$
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
$endgroup$
– greedoid
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 <.
$endgroup$
– greedoid
Dec 19 '18 at 18:34
1
$begingroup$
In second case parabola can touch x-axsis so it can have at most one root
$endgroup$
– greedoid
Dec 19 '18 at 18:34
$begingroup$
I understood this from this. quora.com/…
$endgroup$
– Sherlock Watson
Dec 20 '18 at 2:23
add a comment |
$begingroup$
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
$endgroup$
– greedoid
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 <.
$endgroup$
– greedoid
Dec 19 '18 at 18:34
1
$begingroup$
In second case parabola can touch x-axsis so it can have at most one root
$endgroup$
– greedoid
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
$endgroup$
– greedoid
Dec 19 '18 at 18:33
$begingroup$
Because of > in first and $geq$ second inequality
$endgroup$
– greedoid
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 <.
$endgroup$
– greedoid
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 <.
$endgroup$
– greedoid
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
$endgroup$
– greedoid
Dec 19 '18 at 18:34
$begingroup$
In second case parabola can touch x-axsis so it can have at most one root
$endgroup$
– greedoid
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
add a comment |
$begingroup$
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 $;2a-6ble 36$.
answered Dec 19 '18 at 19:23
BernardBernard
121k740116
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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 $;2a-6ble 36$.
answered Dec 19 '18 at 19:23
BernardBernard
121k740116
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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 $;2a-6ble 36$.
answered Dec 19 '18 at 19:23
BernardBernard
121k740116
$endgroup$
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 $;2a-6ble 36$.
answered Dec 19 '18 at 19:23
BernardBernard
121k740116
edited Dec 20 '18 at 19:19
answered Dec 19 '18 at 19:23
BernardBernard
121k740116
answered Dec 19 '18 at 19:23
BernardBernard
121k740116
answered Dec 19 '18 at 19:23
BernardBernard
121k740116
121k740116
$begingroup$
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.
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– Macavity
Dec 20 '18 at 18:58
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@Macavity: I've added the very simple conclusion for integers.
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– Bernard
Dec 20 '18 at 19:20
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Yes, of course that conclusion works.. +1
$endgroup$
– Macavity
Dec 20 '18 at 19:26
add a comment |
$begingroup$
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
add a comment |
$begingroup$
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.
answered Dec 19 '18 at 18:35
Kshitij MishraKshitij Mishra
413
$endgroup$
add a comment |
$begingroup$
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.
answered Dec 19 '18 at 18:35
Kshitij MishraKshitij Mishra
413
$endgroup$
add a comment |
$begingroup$
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.
answered Dec 19 '18 at 18:35
Kshitij MishraKshitij Mishra
413
$endgroup$
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.
answered Dec 19 '18 at 18:35
Kshitij MishraKshitij Mishra
413
answered Dec 19 '18 at 18:35
Kshitij MishraKshitij Mishra
413
answered Dec 19 '18 at 18:35
Kshitij MishraKshitij Mishra
413
answered Dec 19 '18 at 18:35
Kshitij MishraKshitij Mishra
413
413
add a comment |
add a comment |
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$begingroup$
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