How do you multiply $(10 + \sqrt{20}) - (4 - \sqrt{45})$ $(10 +sqrt20) - (4 - sqrt45)$ ?

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Answer 1 · 2018-05-05T04:14:12

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$(10 + \sqrt{20}) - (4 - \sqrt{45}) = 6 + 5 \sqrt{5}$ $color(red)((10 +sqrt20) - (4 - sqrt45)=6+5 sqrt(5)$

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**Given the radical expression: ** $(10 + \sqrt{20}) - (4 - \sqrt{45})$ $color(blue)((10 +sqrt20) - (4 - sqrt45)$

Remove the brackets:

$\Rightarrow 10 + \sqrt{20} - 4 + \sqrt{45})$ $rArr 10 +sqrt(20) - 4 + sqrt45)$

Rewrite radicals as:

$\Rightarrow 10 + \sqrt{4 \cdot 5} - 4 + \sqrt{9.5}$ $rArr 10+sqrt(4*5)-4+sqrt(9.5$

Use the formula: $\sqrt{a \cdot b} = \sqrt{a} \cdot \sqrt{b}$ $color(blue)(sqrt(a*b)=sqrt(a)*sqrt(b)$

$\Rightarrow 10 + \sqrt{4} \cdot \sqrt{5} - 4 + \sqrt{9} \cdot \sqrt{5}$ $rArr 10+sqrt(4)*sqrt(5)-4+sqrt(9)*sqrt(5$

Simplify after combining the like terms:

$\Rightarrow 10 - 4 + \sqrt{4} \cdot \sqrt{5} + \sqrt{9} \cdot \sqrt{5}$ $rArr 10-4+sqrt(4)*sqrt(5)+sqrt(9)*sqrt(5$

$\Rightarrow 6 + \sqrt{2^{2}} \cdot \sqrt{5} + \sqrt{3^{2}} \cdot \sqrt{5}$ $rArr 6+sqrt(2^2)*sqrt(5)+sqrt(3^2)*sqrt(5$

Use the formula: $\sqrt{a^{2}} = \sqrt{a \cdot a} = a$ $color(blue)(sqrt(a^2)=sqrt(a*a)=a$

$\Rightarrow 6 + 2 \sqrt{5} + 3 \sqrt{5}$ $rArr 6+2sqrt(5)+3sqrt(5)$

Use the formula: $a \sqrt{n} + b \sqrt{n} = (a + b) \sqrt{n}$ $color(blue)(a sqrt(n)+b sqrt(n)=(a+b)sqrt(n)$

$\Rightarrow 6 + (2 + 3) \sqrt{5}$ $rArr 6+(2+3) sqrt(5)$

$\Rightarrow 6 + 5 \sqrt{5}$ $rArr 6+5 sqrt(5)$

Hence,

$(10 + \sqrt{20}) - (4 - \sqrt{45}) = 6 + 5 \sqrt{5}$ $color(blue)((10 +sqrt20) - (4 - sqrt45)=6+5 sqrt(5)$

is the final answer.

How do you multiply (10+√20)−(4−√45)(10 +sqrt20) - (4 - sqrt45)?