Discussion Forum

If $\sqrt{3} = 1.732$, find the value of $\sqrt{196} -\frac{1}{2}\sqrt{48} - \sqrt{75}$
correct to 3 places of decimal.

Answer:

Explanation:

$\sqrt{192} - \frac{1}{2}\sqrt{48} -\sqrt{75}$
$=\sqrt{64\times 3} -\frac{1}{2}\sqrt{16\times 3}-\sqrt{25\times 3}$
= $8\sqrt{3}- \frac{1}{2} \times 4\sqrt{3} -5\sqrt{3}$
= $3\sqrt{3}- 2\sqrt{3}$
$=\sqrt{3} = 1.732.$