Question #69536

Oct 28, 2017

We get, $5 x {y}^{3} \sqrt{6 x z}$ from $\sqrt{150 {x}^{3} {y}^{6} z}$

Explanation:

$5 x {y}^{3} \sqrt{6 x z}$ from $\sqrt{150 {x}^{3} {y}^{6} z}$

Consider
$\sqrt{150 {x}^{3} {y}^{6} z}$

As ${a}^{m} \times {a}^{n} = {a}^{m + n}$,

$\implies \sqrt{\left(25 \times 6\right) \times \left({x}^{2} \times x\right) \times \left({y}^{2} \times {y}^{4}\right) \times z}$

$\implies \sqrt{\left({5}^{2} \times 6\right) \times \left({x}^{2} \times x\right) \times \left({y}^{2} \times {y}^{2} \times {y}^{2}\right) \times z}$

Taking the roots of the square terms outside the radical sign:

$\implies 5 \times \left(x\right) \times y \times y \times y \times \sqrt{6 x z}$

$\implies 5 x {y}^{3} \sqrt{6 x z}$

Oct 28, 2017

$5 x {y}^{3} \sqrt{6 x z}$

Explanation:

$\textcolor{g r e e n}{\text{The trick is to look for squared values}}$

$\textcolor{b l u e}{\text{Dealing with the number part}}$

We have 150.

Just for a moment forget that this is hundreds and just focus on the 15 part. It is known that $3 \times 5 = 15$

Now we deal with the hundreds part. It is known that 15xx10=150. So putting all of this together we have: $3 \times 5 \times 10 = 150$

We know that $2 \times 5 = 10$. Again we can substitute this back in giving: $3 \times 5 \times 2 \times 5 = 150$

Notice that within this we have $5 \times 5$ so we can write:

$3 \times 2 \times {5}^{2}$

Bothe 3 and 2 are prime numbers so there is no way we can square root them. However, $\sqrt{{5}^{2}} = 5$ so we can square root that part.

$\textcolor{red}{\sqrt{150} = 5 \sqrt{3 \times 2} = 5 \sqrt{6}}$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Dealing with the variable part (letters)}}$

${x}^{3} {y}^{6} z$

${x}^{3}$ is the same as ${x}^{2} \times x$
${y}^{6}$ is the same as ${y}^{2 + 2 + 2} = {y}^{2} \times {y}^{2} \times {y}^{2}$

A single $z$ we can do nothing with.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Putting it all together}}$

So we have: $5 \sqrt{6 \times {x}^{2} \times x \times {y}^{2} \times {y}^{2} \times {y}^{2} \times z}$

Taking all the squared values out of the root

$5 x {y}^{3} \sqrt{6 x z}$