# 3root98-root405+root500-root72?

May 23, 2018

$15 \sqrt{2} + \sqrt{5}$

#### Explanation:

$3 \sqrt{98} - \sqrt{405} + \sqrt{500} - \sqrt{72}$
(I take "3root" to mean 3 times the square root, i.e. $3 \sqrt{98}$, and not "cubic root", i.e. $\sqrt[3]{98}$.)

We need to factorise this expression to see if there are any common factors here. We notice that:
$98 = 2 \cdot 49 = 2 \cdot {7}^{2}$
$405 = 9 \cdot 45 = 5 \cdot {9}^{2}$ (9 is a factor is 405 since 4+0+5=9)
$500 = 25 \cdot 20 = {5}^{2} \cdot 5 \cdot 4 = {5}^{3} \cdot {2}^{2} = 5 \cdot {10}^{2}$
$72 = 9 \cdot 8 = 2 \cdot {6}^{2}$

Therefore:
$3 \sqrt{2 \cdot {7}^{2}} - \sqrt{5 \cdot {9}^{2}} + \sqrt{5 \cdot {10}^{2}} - \sqrt{2 \cdot {6}^{2}}$
Move the square numbers outside and take the square root of each:
$3 \cdot 7 \sqrt{2} - 9 \sqrt{5} + 10 \sqrt{5} - 6 \sqrt{2}$
Rearrange so that roots that belong together, come together:
$\left(21 - 6\right) \sqrt{2} + \left(10 - 9\right) \sqrt{5} = 15 \sqrt{2} + \sqrt{5}$

May 23, 2018

$15 \sqrt{2} + \sqrt{5}$

#### Explanation:

$3 \sqrt{98} - \sqrt{405} + \sqrt{500} - \sqrt{72}$

$\therefore = 3 \sqrt{2 \cdot 7 \cdot 7} - \sqrt{5 \cdot 9 \cdot 9} + \sqrt{5 \cdot 10 \cdot 10} - \sqrt{3 \cdot 3 \cdot 2 \cdot 2 \cdot 2}$

$\therefore = 3 \cdot 7 \sqrt{2} - 9 \sqrt{5} + 10 \sqrt{5} - 2 \cdot 3 \sqrt{2}$

$\therefore = 21 \sqrt{2} - 9 \sqrt{5} + 10 \sqrt{5} - 6 \sqrt{2}$

$\therefore = 15 \sqrt{2} + \sqrt{5}$