1. Write the general term

For $\left(\frac{x^{5/2}}{2}-\frac{4}{x^l}\right)^9,$ the general term is $T_{r+1}=\binom{9}{r}\left(\frac{x^{5/2}}{2}\right)^{9-r}\left(-\frac{4}{x^l}\right)^r.$

Simplify: $T_{r+1}=\binom{9}{r}\frac{x^{\frac{5}{2}(9-r)}}{2^{9-r}}\cdot (-1)^r 4^r x^{-lr}.$ Since $4^r=2^{2r}$, $T_{r+1}=\binom{9}{r}(-1)^r 2^{2r-(9-r)}x^{\frac{5}{2}(9-r)-lr}$ $=\binom{9}{r}(-1)^r 2^{3r-9}x^{\frac{45}{2}-r\left(\frac52+l\right)}.$

2. Use the condition for constant term

For constant term, power of $x$ must be zero: $\frac{45}{2}-r\left(\frac52+l\right)=0.$ So $r\left(\frac52+l\right)=\frac{45}{2}.$

Also, its coefficient is given as $-84$.

Now test integer $r\in\{0,1,2,\dots,9\}$ such that the coefficient becomes $-84$.

Coefficient of $T_{r+1}$ is $\binom{9}{r}(-1)^r2^{3r-9}.$

Check odd $r$ (since coefficient is negative):

• $r=1$: $-\binom91 2^{-6}=-\frac{9}{64}$
• $r=3$: $-\binom93 2^0=-84$

So the constant term occurs for $r=3$.

Hence, $3\left(\frac52+l\right)=\frac{45}{2}$ $\frac52+l=\frac{15}{2}$ $l=5.$

3. Find the coefficient of $x^{-3l}$

Since $l=5$, we need the coefficient of $x^{-3l}=x^{-15}.$

Power of $x$ in general term is $\frac{45}{2}-r\left(\frac52+l\right).$ With $l=5$, $\frac52+l=\frac52+5=\frac{15}{2}.$ So exponent becomes $\frac{45}{2}-\frac{15r}{2}.$ Set this equal to $-15$: $\frac{45}{2}-\frac{15r}{2}=-15$ $45-15r=-30$ $-15r=-75$ $r=5.$

Thus required coefficient is coefficient of $T_6$: $\binom95(-1)^5 2^{3\cdot 5-9}$ $=126(-1)2^6$ $=-126\cdot 64.$ Now write it as $2^\alpha\beta$, where $\beta<0$ is odd.

Since $126=2\cdot 63,$ we get $-126\cdot 64=-(2\cdot 63)\cdot 2^6=-63\cdot 2^7.$ So $\alpha=7,\qquad \beta=-63.$

4. Compute $|\alpha l-\beta|$

$|\alpha l-\beta|=|7\cdot 5-(-63)|=|35+63|=98.$

5. Compare with stored answer

Derived answer = $98$. Stored correct answer = $98$. So they agree.