Question

If $\sqrt{28-6\sqrt{3}}$ = $\sqrt{3}a+b$, (where, $a,b$ are rational numbers) find the value of $a-b$.

• A. 4
• B. -4
• C. -3
• D. 2

Answer 1

The correct option is B -4

The expression $\sqrt{28-6\sqrt{3}}$ can be transformed into the identity $(x-y{)}^2$as shown below,
$\sqrt{28-6\sqrt{3}}$ = $\sqrt{(1{)}^2-2\times 1\times 3\sqrt{3}+(3\sqrt{3}{)}^2}$
= $\sqrt{(1-3\sqrt{3}{)}^2}$
= $1-3\sqrt{3}$
Now, the given equation can be written as,
$1-3\sqrt{3}$ = $\sqrt{3}a+b$
Comparing the LHS to RHS, we get,
$a=-3$ and $b=1$
Thus, $a-b=-3-1=-4$