Mathematics - Quadratic Equation Question with Solution | TestHub

Given,

${\log }_2\left(9^{2\alpha -4}+13\right)-{\log }_2\left(3^{2\alpha -4}·\frac{5}{2}+1\right)=2$

Now let $3^{2\alpha -4}=t$, so the equation becomes,

${\log }_2\left(t^2+13\right)-{\log }_2\left(\frac{5t}{2}+1\right)=2$

$\Rightarrow {\log }_2\frac{\left(t^2+13\right)}{\left({\displaystyle \frac{5t}{2}}+1\right)}=2$

$\Rightarrow \frac{\left(t^2+13\right)}{\left({\displaystyle \frac{5t}{2}+1}\right)}=2^2$

$\Rightarrow t^2+13=10t+4$

$\Rightarrow t^2-10t+9=0$

$\Rightarrow t=1 \text{or} 9$

So,

$3^{2\alpha -4}=1 \text{or} 9$

$\Rightarrow 3^{2\alpha -4}=3^0 \text{or} 3^2$

$\Rightarrow 2\alpha -4=0 \mathrm{or} 2$

$\Rightarrow \alpha =2, 3$

Now,

${\mathrm{x}}^2-2{\left(\sum _{\alpha \in s}\alpha \right)}^2\mathrm{x}+\sum _{a\in s}{\left(\alpha +1\right)}^2\beta =0$

$\Rightarrow x^2-2\left({\left(2+3\right)}^{2}x\right)+\left(3^2+4^2\right)\beta =0$

$\Rightarrow x^2-50x+25\beta =0$

Now for real roots 

$D\ge 0$

$\Rightarrow {50}^2-4\times 25\beta \ge 0$

$\Rightarrow 50-2\beta \ge 0$

$\Rightarrow \beta \le 25$

So, maximum value of $\beta$ is $25$.