Question

If $\sqrt{11-3\sqrt{8}}=a+b\sqrt{2}$, then what is the value of (2a+3b)?

• A. 7
• B. 9
• C. 3
• D. 5

Answer 1

Answer:

C. 3

Solution:

Concept used:

(a – b)² = a² – 2ab + b² 

Calculations:

$\sqrt{11-3\sqrt{8}}=a+b\sqrt{2}$

⇒$\sqrt{11-3\sqrt{2\times{2}\times{2}}}=a+b\sqrt{2}$

⇒$\sqrt{11-2\times{3\sqrt{2}}}=a+b\sqrt{2}$

⇒$\sqrt{(3)^2+(\sqrt{2})^2-2\times{3\sqrt{2}}}=a+b\sqrt{2}$

⇒$\sqrt{(3-\sqrt{2})^2}=a+b\sqrt{2}$

⇒ $3-2\sqrt{2} = a + b\sqrt{2}$

Compare a and b 

⇒ a = 3 

⇒ b = -1 

Value of (2a + 3b) = $2\times{3}+3\times{(-1)}$ 

⇒ 6 – 3 = 3 

∴ Value of 2a + 3b is 3