CAT 2024

Slot 1

QA

Question & Solution

$(a + b \sqrt{n})$ is the positive square root of $(29 - 12\sqrt{5})$

So $29-12\sqrt{5}=\left(a+b\sqrt{n}\right)^2$
$29-12\sqrt{5}=a^2+b^2n+2ab\sqrt{n}$

$a^2+b^2n=29$ and
$ab\sqrt{n}=-6\sqrt{5}$
$a^2b^2n=180$
$b^2n=\frac{180}{a^2}$
Substituting this in the above equation, 
$a^2+\frac{180}{a^2}=29$
$a^4-29a^2+180=0$
$a^2=\frac{\left(29\pm\sqrt{29^2-4\left(180\right)}\right)}{2}$
$a^2=\frac{\left(29+\sqrt{841-720}\right)}{2}$
$a^2=9\ or\ 20$

That means, one of $a^2\ or\ b^2n$ is 9 and 20. 

We also have, $ab\sqrt{n}=-6\sqrt{5}$ that means one of a or b should be negative
And also the fact that this is a positive square root,
And we need to maximise the value of a, b and n. 

We can have a=-3, b=1 and n=20. 
This satisfies all the above equations, and the value of a+b+n=18.