Question 1192821
<font color=black size=3>
Notice that 128 = 64*2 showing that 64 is the largest perfect square factor of 128.
Similarly, x^2 is the largest perfect square factor of x^3


We can rewrite 128x^3 into 64x^2*2x to be able to perform this simplification
{{{sqrt(128x^3) = sqrt(64x^2*2x)}}}
{{{sqrt(128x^3) = sqrt(64x^2)*sqrt(2x)}}} Use the rule that sqrt(A*B) = sqrt(A)*sqrt(B)
{{{sqrt(128x^3) = sqrt((8x)^2)*sqrt(2x)}}}
{{{sqrt(128x^3) = 8x*sqrt(2x)}}}
Such that {{{x >= 0}}}


Through similar logic, we can also say:
{{{sqrt(98x^2) = sqrt(49x^2*2)}}}
{{{sqrt(98x^2) = sqrt(49x^2)*sqrt(2)}}}
{{{sqrt(98x^2) = sqrt((7x)^2)*sqrt(2)}}}
{{{sqrt(98x^2) = 7x*sqrt(2)}}}


And also,
{{{sqrt(162x) = sqrt(81*2x)}}}
{{{sqrt(162x) = sqrt(81)*sqrt(2x)}}}
{{{sqrt(162x) = 9*sqrt(2x)}}}


Use the three results we got earlier above to simplify the given expresdsion like so
{{{5*sqrt(128x^3)+4*sqrt(98x^2)-3x*sqrt(162x)}}}


{{{5*8x*sqrt(2x)+4*7x*sqrt(2)-3x*9*sqrt(2x)}}}


{{{40x*sqrt(2x)+28x*sqrt(2)-27x*sqrt(2x)}}}


{{{40x*sqrt(2x)-27x*sqrt(2x)+28x*sqrt(2)}}}


{{{(40x-27x)*sqrt(2x)+28x*sqrt(2)}}}


{{{13x*sqrt(2x)+28x*sqrt(2)}}}


Therefore,
{{{5*sqrt(128x^3)+4*sqrt(98x^2)-3x*sqrt(162x)}}} simplifies to {{{13x*sqrt(2x)+28x*sqrt(2)}}}


Or we can say
{{{5*sqrt(128x^3)+4*sqrt(98x^2)-3x*sqrt(162x)=13x*sqrt(2x)+28x*sqrt(2)}}}
only when {{{x >= 0}}}
</font>