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 $x$, then the largest possible value of $2a−6b$ is

Question

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?

Answer 1

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

So $$ a\leq 3\;\;\;\;{\rm and}\;\;\;\;b\geq -5 \implies 2a-6b\leq 36$$

Answer 2

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.

Answer 3

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 $p\Bigl(-\dfrac B{2A}\Bigr)>0\:$ (resp. $\ge 0$).

Here the hypotheses yield \begin{cases}2\Bigl(\dfrac a4\Bigr)^2-\dfrac{a^2}4+2>\iff2-\dfrac{a^2}8>0\iff a^2<16, \\[1ex] \Bigl(\dfrac b2\Bigr)^2-\dfrac{b^2}2+8\ge 0\iff 8-\dfrac{b^2}4\ge 0\iff b^2\le 32, \end{cases} Therefore we have $-4<a<4\;$ and $\;-4\sqrt 2 \le b\le 4\sqrt 2$, so that $$\left.\begin{matrix}-8<2a<8\\-24\sqrt 2\le -6b\le24\sqrt2\end{matrix}\right\}\Rightarrow -8(1+3\sqrt2)<2a-6b<8(1+3\sqrt 2).$$

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 $$\begin{cases} a^2<16\\b^2\le 32\end{cases}\iff \begin{cases} |a|<4\\ |b|\le 5\end{cases}\iff \begin{cases} -3\le a \le 3\\ -5\le b\le 5\end{cases}\iff \begin{cases} -6\le 2a \le 6\\ -30\le -6b\le 30\end{cases}$$ so $\;2a-6b\le 36$.