## anonymous 4 years ago √45k^7q^8

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1. Mertsj

$\sqrt{9(5)k ^{6}kq ^{8}}$

2. Mertsj

Rewrite it like that and then take the square root of all the perfect square factors.

3. vishweshshrimali5

$\large\sqrt{45 k^7 q^8}$ Now , $\large k^7 = \cfrac{k^7 * \color {red}k}{\color {green}k} = \cfrac{k^8}{k}$ also, $\large\sqrt{k^8} = k^4$ and $\large \sqrt{q^8} = q^4$ and $\large \sqrt{45} = \sqrt{9*5} = \sqrt{9}\sqrt{5} = 3\sqrt{5}$ Now you can put all the values in the question $\large\sqrt{45 k^7 q^8}$ $\large \color {green}{\sqrt{45}} * \color {red}{\sqrt{k^7}} * \color {brown}{\sqrt{q^8}}$ we have already calculated $\large \sqrt{45}$ $\large \sqrt{q^8}$ and how to change $$\large k^7$$ into $$\large \cfrac{k^8}{k}$$ also we have find out $\large\sqrt{k^8}$ You can solve it further also the way shown by @Mertsj is a lot easier you can use either of them.

4. vishweshshrimali5

@Mertsj am I wrong somewhere ...... ?

5. Mertsj

$3k^3q^4\sqrt{5k}$

6. vishweshshrimali5

$\large \cfrac{3k^4 q^4 \sqrt{5}}{\sqrt{k}}$ $\large \cfrac {3k^4 q^4 \sqrt{5}*\sqrt{k}}{\sqrt{k}*\sqrt{k}}$ $\large 3k^3 q^4 \sqrt{5k}$ Yours method is easier @Mertsj gud work