Question

Express in simplest radical form. 5x√192x³+√3x5 please help!!

Answer 1

$-5x\sqrt{192x^3} + \sqrt{3x^5}\\\\=-5x\left(192x^3 \right)^{\tfrac 12} + \left(3x^5 \right)^{\tfrac 12}\\\\=-5\cdot (192)^{\tfrac 12} \cdot x \cdot x^{\tfrac 32} + 3^{\tfrac 12} \cdot x^{\tfrac 52}\\\\=-5\left( 64 \cdot 3 \right)^{\tfrac 12} \cdot x^{1+\tfrac 32} + 3^{\tfrac 12} \cdot x^{\tfrac 52}\\\\=-5\cdot 8\cdot 3^{\tfrac 12} \cdot x^{\tfrac 15} + 3^{\tfrac 12} \cdot x^{\tfrac 52}\\\\=-40\cdot 3^{\tfrac 12} \cdot x^{\tfrac 52}+3^{\tfrac 12} \cdot x^{\tfrac 52}$

$=-39\cdot 3^{\tfrac 12} \cdot x^{\tfrac 52}\\\\=-39\sqrt 3 \sqrt{x^5}$