Question

If square root of 17 +12 root 2/ sq root of 11 + root 72 = a+b root 2, then what is the value of (a+3b)?
a 2 b 1 2/7 c 1 d 1 3/7

Answer 1

Answer:

2

Step-by-step explanation:

Answer 2

Given:$\[\frac{{\sqrt {17 + 12\sqrt 2 } }}{{\sqrt {11 + \sqrt {72} } }} = a + b\sqrt 2 \]$

To find: the value of $\[\left( {a + 3b} \right)\]$.

Solution:

Take the numerator of the LHS and simplify it.

$\[\sqrt {17 + 12\sqrt 2 } = \sqrt {9 + 8 + 2 \times 3 \times 2\sqrt 2 } \]$

                    $\[ = \sqrt {{3^2} + {{\left( {2\sqrt 2 } \right)}^2} + 2 \times 3 \times 2\sqrt 2 } \]$

$\[[\]$∵ the above expression is of the form $\[{\left( {a + b} \right)^2}]\]$

                    $\[\begin{array}{l} = \sqrt {{{\left( {3 + 2\sqrt 2 } \right)}^2}} \\ = 3 + 2\sqrt 2 \end{array}\]$

Take denominator of the RHS and simplify it.

$\[\sqrt {11 + \sqrt {72} } = \sqrt {11 + \sqrt {36 \times 2} } \]$

                  $\[\begin{array}{l} = \sqrt {11 + 6\sqrt 2 } \\ = \sqrt {9 + 2 + 3 \times 2\sqrt 2 } \\ = \sqrt {{3^2} + {{\left( {\sqrt 2 } \right)}^2} + 2 \times 3 \times \sqrt 2 } \end{array}\]$

$\[[\]$∵ the above expression is of the form $\[{\left( {a + b} \right)^2}]\]$

                 $\[\begin{array}{l} = \sqrt {{{\left( {3 + \sqrt 2 } \right)}^2}} \\ = 3 + \sqrt 2 \end{array}\]$

Observe that,

$\[\frac{{3 + 2\sqrt 2 }}{{3 + \sqrt 2 }} = \frac{{3 + \sqrt 2 + \sqrt 2 }}{{3 + \sqrt 2 }}\]$

          $\[ = \frac{{3 + \sqrt 2 }}{{3 + \sqrt 2 }} + \frac{{\sqrt 2 }}{{3 + \sqrt 2 }}\]$

          $\[ = 1 + \frac{{\sqrt 2 }}{{3 + \sqrt 2 }} \times \frac{{3 - \sqrt 2 }}{{3 - \sqrt 2 }}\]$

          $\[ = \frac{{\left( {9 - 2} \right) + \sqrt 2 \left( {3 - \sqrt 2 } \right)}}{{\left( {9 - 2} \right)}}\]$

          $\[ = \frac{{7 + 3\sqrt 2 - 2}}{7}\]$

          $\[ = \frac{{5 + 3\sqrt 2 }}{7}\]$

          $\[ = \frac{5}{7} + \frac{3}{7}\sqrt 2 \]$

Compare $\[ \frac{5}{7} + \frac{3}{7}\sqrt 2 \]$ with $\[a + b\sqrt 2 \]$.

Understand that, $\[a = \frac{5}{7},b = \frac{3}{7}\]$

Find the value of $\[\left( {a + 3b} \right)\]$.

$\[\begin{array}{l}\left( {a + 3b} \right) = \frac{5}{7} + 3\left( {\frac{3}{7}} \right)\\ \Rightarrow \left( {a + 3b} \right) = \frac{5}{7} + \frac{9}{7}\\ \Rightarrow \left( {a + 3b} \right) = \frac{{5 + 9}}{7}\end{array}\]$

$\[\begin{array}{l} \Rightarrow \left( {a + 3b} \right) = \frac{{14}}{7}\\ \Rightarrow \left( {a + 3b} \right) = 2\end{array}\]$

Hence, the value of $\[\left( {a + 3b} \right)\]$ is 2.