Applying binomial expansion and using algebra to find the value of $a+b$

Question

Q) In the expansion of $f(x)=(1 + ax)^4 (1 + bx)^5$ where $a$ and $b$ are positive integers, the coefficient of $x^2$ is 66. Evaluate $a+b$.

My working:

After expanding the expression I simplified it and got

$5b^2+10ab+3a^2=33$

After further simplification, I managed to get

$5(a+b)^2-2a^2=33$

but I'm not sure what to do next.

I agree with your working, and I'm also not sure what to do next. Hmm.... — Adam Rubinson, Sep 05 '21 at 17:06
Note that $1\leq a\leq a+b$ so $3a^2 \leq 5(a+b)^2-2a^2\leq 33$. Thus $1\leq a\leq 3$. Now, I assume you can check the three values of $a$. — Kapil, Sep 05 '21 at 17:11
Oh, they're positive integers. That narrows it down. I didn't read that part properly the first time round. — Adam Rubinson, Sep 05 '21 at 17:17
@Kapil according to your way , no integral values satisfy this . — Sukhoi234, Sep 05 '21 at 17:25

lone student · Answer 1 · 2021-09-15T06:17:54.717

3

Your result is correct.

If $a,b\in\mathbb Z^{+}$ then we can write,

$$\begin{align}&5b^2+3a^2≥2\sqrt {15}ab\\ \implies &5b^2+3a^2+10ab≥ab \left(10+2\sqrt {15}\right)\\ \implies &33≥\left(10+2\sqrt {15}\right)ab\\ \implies &ab≤\frac{33}{10+2\sqrt {15}}<2\\ \implies &a=b=1.\end{align}$$

This means, the solution doesn't exist.

edited Sep 15 '21 at 06:17

answered Sep 05 '21 at 17:29

lone student

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Applying binomial expansion and using algebra to find the value of $a+b$

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