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# help math

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Calculate sqrt(60x)*sqrt(12x)*sqrt(63x)*sqrt(42x). Express your answer in simplest radical form in terms of x. Note: When entering a square root with more than one character, you must use parentheses or brackets. For example, you should enter sqrt(14) as "sqrt(14)" or "sqrt{14}".

Jul 3, 2021

#1
+171
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oh lord, this is hard to swallow

oh well, lets start shall we -- pay attention to every step:

$\Huge\text{NOTE:}$  this website kind of screwed up the colouring i do not know why but check this link for a more accurate version: http://mathb.in/59340

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$\color{lightskyblue}{\sqrt{60x}} \sqrt{12x}\sqrt{63x}\sqrt{42x}$

$\color{lightskyblue}{\sqrt{4(15)x}}\sqrt{12x}\sqrt{63x}\sqrt{42x}$

$\color{lightskyblue}{ \sqrt{2^2(15)x}}\sqrt{12x}\sqrt{63x}\sqrt{42x}$

$\color{lightskyblue}{2\sqrt{(15)x}}\sqrt{12x}\sqrt{63x}\sqrt{42x}$

now lets do the second one: $2\sqrt{(15)x} \color{tomato}{\sqrt{12x}}\sqrt{63x}\sqrt{42x}$

$2\sqrt{(15)x} \color{tomato}{ \sqrt{4(3)x}}\sqrt{63x}\sqrt{42x}$

$2\sqrt{(15)x} \color{tomato}{ \sqrt{2^2(3)x}}\sqrt{63x}\sqrt{42x}$

$2\sqrt{(15)x} \color{tomato}{ 2 \sqrt{(3)x}}\sqrt{63x}\sqrt{42x}$

now lets do the third one: $2\sqrt{(15)x}2 \sqrt{(3)x} \color{teal}{\sqrt{63x}} \sqrt{42x}$

$2\sqrt{(15)x}2 \sqrt{(3)x} \color{teal}{\sqrt{9(7)x}} \sqrt{42x}$

$2\sqrt{(15)x}2 \sqrt{(3)x} \color{teal}{\sqrt{3^3(7)x}} \sqrt{42x}$

$2\sqrt{(15)x}2 \sqrt{(3)x} \color{teal}{(3\sqrt{(7)x}) } \sqrt{42x}$

We are left with

$2\sqrt{(15)x}2 \sqrt{(3)x} (3\sqrt{(7)x}) \sqrt{42x}$

firstly lets multiply the first two terms:

$\color{darkorange}{2\sqrt{(15)x}2 \sqrt{(3)x}} (3\sqrt{(7)x}) \sqrt{42x}$

$\color{darkorange}{4\sqrt{(15)x} \sqrt{(3)x}} (3\sqrt{(7)x}) \sqrt{42x}$

$\color{darkorange}{4\sqrt{3x(15x)}} (3\sqrt{(7)x}) \sqrt{42x}$

$\color{darkorange}{4\sqrt{45x^2}} (3\sqrt{(7)x}) \sqrt{42x}$

$\color{darkorange}{4\sqrt{9(5)x^2}} (3\sqrt{(7)x}) \sqrt{42x}$

$\color{darkorange}{4\sqrt{3^2(5)x^2}} (3\sqrt{(7)x}) \sqrt{42x} \Leftrightarrow \color{darkorange}{4\sqrt{3^2 \cdot x^2(5)}} (3\sqrt{(7)x}) \sqrt{42x}$

$\color{darkorange}{4\sqrt{(3x)^2(5)}} (3\sqrt{(7)x}) \sqrt{42x}$

$\color{darkorange}{4[(3x)\sqrt{5}]} (3\sqrt{(7)x}) \sqrt{42x}$

$\color{darkorange}{12x\sqrt{5}} (3\sqrt{(7)x}) \sqrt{42x}$

now lets do work the $12x\sqrt{5} (3\sqrt{(7)x})$ :

$\color{mediumseagreen}{12x\sqrt{5} (3\sqrt{(7)x})} \sqrt{42x}$

$\color{mediumseagreen}{36x\sqrt{7x\cdot 5} } \sqrt{42x}$

$\color{mediumseagreen}{36x\sqrt{35x} } \sqrt{42x}$

finally, we multiply them all together:

$\color{navy}{36x\sqrt{35x} \sqrt{42x}}$

$\color{navy}{36x\sqrt{42x(35x)}}$

$\color{navy}{36x\sqrt{1470x^2}}$

$\color{navy}{36x\sqrt{(7x)^2 \cdot 30 }}$

$\color{navy}{36x(7x\sqrt{30})}$

$\color{navy}{36x\cdot x(7\sqrt{30})}$

$\color{navy}{36x^2(7\sqrt{30})}$

$\color{navy}{252x^2(\sqrt{30})}$

or just

$\color{navy}{252x^2\sqrt{30}}$

$\color{navy}{\left(6\sqrt{7}\right)x^2\sqrt{30}}$

$\color{navy}{6\sqrt{7\cdot \:30}x^2}$

$\boxed{\color{navy}{ 6\sqrt{210}x^2}}$

can you do anything else to this? i dont think so.

Jul 3, 2021