Question

If $19-6\sqrt{10}=a^2,$ find the value of $a.$

• A. $\sqrt{10}-3$
• B. $\sqrt{3}-10$
• C. $\sqrt{10}+3$
• D. Cannot be determined.

Answer 1

The correct option is A $\sqrt{10}-3$

We know, $(x-y{)}^2=x^2+y^2-2xy$

Let's express $19-6\sqrt{10}$ in $x^2+y^2-2xy$ form, then determine $(x-y{)}^2$.

$19-6\sqrt{10}$
$=10+9-2\times 3\times \sqrt{10}$
$=(\sqrt{10}{)}^2+(\sqrt{9}{)}^2-2\times \sqrt{9}\times \sqrt{10}$
$=(\sqrt{10}-\sqrt{9}{)}^2$
$=(\sqrt{10}-3{)}^2$

Comparing with $19-6\sqrt{10}=a^2,a$ is $\sqrt{10}-3.$