Calculation of all values of $a$ for which $3x^2+(4-2a)x-8-a^2\leq 0$.
Given that $x$ lies between $-3$ and $2$.
My Try:: Let $x = \alpha,\beta$ be the Roots of Given equation. where $-3<\alpha,\beta<2$
So $\displaystyle \alpha+\beta = \frac{2a-4}{3}$ and $\displaystyle \alpha.\beta = \frac{-(8+a^2)}{3}$
Now I Did not Understand How can I proceed
Help Required
Thanks
The inequality can be written as:
$$(2x+1)^2 -9 \le (a+x)^2$$
Now if $-3 \le x \le 2$, the LHS takes values of $16$ at the end points. For the inequality to hold, we must then have $16 \le (a+2)^2$ and $16 \le(a-3)^2$. As the RHS is less convex, or is a flatter parabola, the intermediate points will fall in line.
Solving these, we have $a \le -6$ or $a \ge 2$ from the first condition
and $a \le -1$ or $a \ge 7$ from the second. Hence we can have any real $a \not \in (-6,7)$.
Hint : What you have done is correct. But, don't you think it would be simple to solve the quadratic equation; this will give you the two roots of the equations as a function of parameter "a". Then apply the conditions. Are you able to continue with this ?
