can someone help me explain these questions
\[27. -2\sqrt{7x ^{2}}*\frac{ 1 }{ 3 } \sqrt{28x ^{3}}\]
\[33. (3\sqrt{2}-2\sqrt{5})(4\sqrt{2}+ 2\sqrt{5})\]
\[43. n \sqrt{2}=\sqrt{9-3n}\]
\[47. y=\sqrt{x+4}\]
please help explain it
@zepdrix
\[\Large\bf\sf \color{royalblue}{27)} -2\sqrt{7x ^{2}}\cdot\frac{1}{3}\sqrt{28x ^{3}}\]Multiplication is commutative, so we can multiply things in any order. Let's bring the 1/3 to the front and multiply it with the -2.\[\Large\bf\sf \color{royalblue}{27)} -\frac{2}{3}\sqrt{7x ^{2}}\cdot\sqrt{28x ^{3}}\]Then we'll combine our square roots by multiplying the insides together,\[\Large\bf\sf \color{royalblue}{27)} -\frac{2}{3}\sqrt{7x^2\cdot 28x ^{3}}\]
From there we can multiply some stuff together under the root. Following along so far?
yes
@zepdrix
So again, under the root we're multiplying so we can move stuff around. We'll multiply the 7 and 28. And then we'll multiply the x^2 by x^3. What do the x's give you?
\[\Large\bf\sf x^2\cdot x^3\quad=\quad ?\]
\[x^{6}\]
No silly! When we multiply terms of similar bases, we `add the exponents`.\[\Large\bf\sf x^2\cdot x^3\quad=\quad x^{2+3}\quad=\quad x^5\]
\[\Large\bf\sf \color{royalblue}{27)} -\frac{2}{3}\sqrt{7\cdot28x^5}\]
oo sorry
Multiplying the 7 and 28, \[\Large\bf\sf \color{royalblue}{27)} -\frac{2}{3}\sqrt{196x^5}\]Now we need to ask ourselves, does 196 have any factors that are perfect squares?
Oh nevermind, 196 is itself a perfect square it seems! :)
ok
What's the square root of 196?
its 14
Ok good so we'll take the square root of 196, \[\Large\bf\sf \color{royalblue}{27)} -\frac{2}{3}\cdot 14\sqrt{x^5}\]Then multiply the 14 by the -2/3, do you understand how to do that?
no
You can think of \(\Large\bf\sf 14\) as \(\Large\bf\sf \dfrac{14}{1}\), and then just do fraction multiplication, \[\Large\bf\sf \color{royalblue}{27)} -\frac{2}{3}\cdot \frac{14}{1}\sqrt{x^5}\] top with top, bottom with bottom,\[\Large\bf\sf \color{royalblue}{27)} -\frac{2\cdot14}{3\cdot1}\sqrt{x^5}\]
ok
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