Question

If $y = 3 + 2\sqrt{2}$, then find the value of $\sqrt{y}$.

• A. $\sqrt2+1$
• B. $1$
• C. $\sqrt2-1$
• D. $\sqrt2+\sqrt3$

Answer 1

Answer:

$\sqrt2+1$

The value of $\sqrt{y}$ is $\sqrt{2} + 1$.

1. Given equation:
   $y = 3 + 2\sqrt{2}$

2. Rewrite the number 3 to match the algebraic identity (a+b)² = a² + 2ab + b²:
   We can split 3 into 2 + 1:
   $y = 2 + 1 + 2\sqrt{2}$
   $y = 2 + 2\sqrt{2} + 1$

3. Express the terms as perfect squares:

   • $2 = (\sqrt{2})^2$

   • 1 = (1)²

   • $2\sqrt{2} = 2 \times \sqrt{2} \times 1$

   Substituting these back into the equation:
   $y = (\sqrt{2})^2 + 2(\sqrt{2})(1) + (1)^2$

4. Simplify using the identity (a+b)²:
   $y = (\sqrt{2} + 1)^2$

5. Take the square root on both sides:
   $\sqrt{y} = \sqrt{(\sqrt{2} + 1)^2}$
   $\mathbf{\sqrt{y} = \sqrt{2} + 1}$