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

Given that,

• $2x^{2} – ax + 2 > 0 \quad \longrightarrow (1)$
• $x^{2} – bx + 8 \geqslant 0 \quad \longrightarrow (2)$

For any quadratic equation $ax^{2} + bx + c > 0$
If $a$ is greater than zero then this means that the quadratic equation will always be above $x\text{-axis}$ and will never intersect it at any real value of $x.$ Thus the solutions to this equation will be imaginary.

So, here discriminant $\boxed{\text{D<0}}$

$\Rightarrow \boxed{ b^{2} – 4ac < 0}$

On the other hand for an inequality $ax^{2} + bx + c < 0$ for $a < 0$ the expression will always be below the $x\text{-axis}.$Similarly, the solutions will be imaginary.

So, here discriminant $\boxed{D<0}$

$\Rightarrow \boxed{b^{2} – 4ac}$

For the equation $(1),$ graph will be :

Here, $\boxed{\text{D}<0}$

$\Rightarrow \boxed{ b^{2} – 4ac < 0}$

$\Rightarrow (-a)^{2} – 4(2)(2) < 0$

$\Rightarrow a^{2} – 16 < 0$

$\Rightarrow a^{2} < 16$

$\Rightarrow \boxed{ -4 < a < 4}$

For the equation $(2),$ graph will be

In this case, the graph can touch $x – \text{axis}$, so it can have at most one root.

Thus discriminant $\boxed{\text{D} \leqslant 0}$

$\Rightarrow \boxed{ b^{2} – 4ac \leqslant 0}$

$\Rightarrow (-b)^{2} – 4(1)(8) \leq 0$

$\Rightarrow b^{2} – 32 \leq 0$

$\Rightarrow b^{2} \leq 32$

$\Rightarrow b \leq \sqrt{32}$

$\Rightarrow b \leq \sqrt{16 \times 2}$

$\Rightarrow \boxed{-4\sqrt{2} \leqslant b \leqslant 4 \sqrt{2}}$

As $b$ is an integer.

So, $\boxed{ -5 \leqslant b \leqslant 5}$

For the largest value of $2a – 6b,$ we can take $a = 3,$ and $b = -5.$

$\therefore$ The largest possible value of $2a – 6b = 2(3) – 6(-5) = 6 + 30 = 36.$

Correct Answer $:36$

$\textbf{PS :}$ For equation $ax^{2} + bx + c = 0,$ expression $b^{2} – 4ac$ is called discriminant and denoted by $\text{D}.$

$\begin{array}{cl|l}\hline & \textbf{Value of discrimimant} & \textbf{Nature of roots}\\\hline 1. & D<0 & \text{Unequal and imaginary}\\ 2. & D = 0 & \text{Real and equal}\\ 3. & D>0, \;\text{and is a perfect square} & \text{Real, unequal, and rational}\\ 4. & D>0,\;\text{and not a perfect square} & \text{Real. unequal, and irrational}\\\hline \end{array}$

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