Algebra 1

Published by Prentice Hall
ISBN 10: 0133500403
ISBN 13: 978-0-13350-040-0

Chapter 10 - Radical Expressions and Equations - Mid-Chapter Quiz - Page 619: 18

Answer

$x\sqrt6$

Work Step by Step

$\frac{-3\sqrt{14x^3}}{-\sqrt{21x}}\longrightarrow$ factor the radicands; use mental math to see that 7x is a factor in both radicands =$\frac{-3\sqrt{2\times{x^2}\times7x}}{-\sqrt{3\times{7x}}}\longrightarrow$ multiplication property of square roots =$\frac{-3\times\sqrt2\times\sqrt{x^2}\times\sqrt{7x}}{-\sqrt3\times\sqrt{7x}}\longrightarrow$ simplify; $\sqrt{7x}$ cancels; negative cancels =$\frac{-3\times\sqrt2\times x}{-\sqrt3}\longrightarrow$ simplify =$\frac{3x\sqrt2}{\sqrt3}\longrightarrow$ multiply numerator and denominator by $\sqrt3$ =$\frac{3x\sqrt2}{\sqrt3}\times\frac{\sqrt3}{\sqrt3}\longrightarrow$ simplify; multiplication property of square roots =$\frac{3x\sqrt6}{3}\longrightarrow$ simplify; 3 cancels =$x\sqrt6$
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