Question

g. $-3sqrt{20}+sqrt{5}$
i. $4sqrt{6}*3sqrt{12}$

Explanation:

Part g: $-3\sqrt{20} + \sqrt{5}$

Step1: Simplify $\sqrt{20}$

We know that $20 = 4\times5$, so $\sqrt{20}=\sqrt{4\times5}$. By the property of square roots $\sqrt{ab}=\sqrt{a}\times\sqrt{b}$ (where $a = 4$, $b = 5$ and $a\geq0,b\geq0$), we have $\sqrt{4\times5}=\sqrt{4}\times\sqrt{5}=2\sqrt{5}$.

Step2: Substitute back and simplify

Substitute $\sqrt{20} = 2\sqrt{5}$ into the original expression: $-3\sqrt{20}+\sqrt{5}=-3\times(2\sqrt{5})+\sqrt{5}$.
First, calculate $-3\times(2\sqrt{5})=-6\sqrt{5}$. Then the expression becomes $-6\sqrt{5}+\sqrt{5}$.
Combine like terms: $(-6 + 1)\sqrt{5}=-5\sqrt{5}$.

Step1: Multiply the coefficients and the square roots separately

Using the property of multiplication of square roots $\sqrt{a}\times\sqrt{b}=\sqrt{ab}$ (where $a\geq0,b\geq0$) and the commutative property of multiplication, we can rewrite the expression as $(4\times3)\times(\sqrt{6}\times\sqrt{12})$.
Calculate $4\times3 = 12$. And $\sqrt{6}\times\sqrt{12}=\sqrt{6\times12}=\sqrt{72}$.

Step2: Simplify $\sqrt{72}$

We know that $72=36\times2$, so $\sqrt{72}=\sqrt{36\times2}=\sqrt{36}\times\sqrt{2}=6\sqrt{2}$ (since $\sqrt{36} = 6$).

Step3: Multiply the results

Now we have $12\times\sqrt{72}=12\times6\sqrt{2}$. Calculate $12\times6 = 72$. So the result is $72\sqrt{2}$.

Answer:

$-5\sqrt{5}$

Part i: $4\sqrt{6}\times3\sqrt{12}$