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Q.
If $x$ is real, then the maximum value of $3-6 x-8 x^{2}$ is
Complex Numbers and Quadratic Equations
Solution:
Let $y=3-6 x-8 x^{2}$
then $8 x^{2}+6 x+y-3=0$.
Since $x$ is real,
$\therefore 6^{2}-4 \cdot 8(y-3) \geq 0, $
or $ 36-32 y+96 \geq 0$
or $32 y \leq 132$
$\therefore y \leq \frac{132}{32}$
or $y \leq \frac{33}{8}$
Hence, maximum value of $y=\frac{33}{8}$.