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Q.
If the minimum value of $f(x)=2 x^{2}+\alpha x+8$ is the same as the maximum value of $g(x)=-3 x^{2}-4 x+\alpha^{2}$, then $\alpha^{2}=$
TS EAMCET 2019
Solution:
Since, the minimum value of
$f(x)=2 x^{2}+\alpha x+8$ is $-\frac{\alpha^{2}-64}{8}=\frac{64-\alpha^{2}}{8}$
And the maximum value of $g(x)=-3 x^{2}-4 x+\alpha^{2}$
is $-\frac{12 \alpha^{2}-16}{-12}=\frac{16+12 \alpha^{2}}{12}$
Now, according to the question
$\frac{64-\alpha^{2}}{8}=\frac{16+12 \alpha^{2}}{12}$
$\Rightarrow 192-3 \alpha^{2}=32+24 \alpha^{2}$
$\Rightarrow 27 \alpha^{2}=160$
$ \Rightarrow \alpha^{2}=\frac{160}{27}$