Chapter 6.4, Problem 21WE

To determine

To Simplify:

  ${\left(2\sqrt{5}+\sqrt{7}\right)}^2$

Answer to Problem 21WE

  $27+4\sqrt{35}$

Explanation of Solution

Given:

The expression is :

  ${\left(2\sqrt{5}+\sqrt{7}\right)}^2$

Formula Used:

  ${\left(a+b\right)}^2=a^2+2ab+b^2$

Calculation:

Given:

  ${\left(2\sqrt{5}+\sqrt{7}\right)}^2\mathrm{..............}(1)$

Use

  ${\left(a+b\right)}^2=a^2+2ab+b^2\mathrm{..........}(2)$

Comparing equation (1) with (2),

  $a=2\sqrt{5},b=\sqrt{7}$

Substitute the values in equation (2),

  $\begin{array}{l}{\left(a+b\right)}^2\\ =a^2+2ab+b^2\\ \Rightarrow {\left(2\sqrt{5}+\sqrt{7}\right)}^2\\ ={\left(2\sqrt{5}\right)}^2+2\left(2\sqrt{5}\right)\left(\sqrt{7}\right)+{\left(\sqrt{7}\right)}^2\\ =2^2\cdot 5+4\sqrt{5\cdot 7}+\mathrm{7.................}[\text{Product property of radicals;}\sqrt[n]{a}\cdot \sqrt[n]{b}=\sqrt[n]{ab}]\\ =20+4\sqrt{35}+7\\ =27+4\sqrt{35}\end{array}$