Chapter 11.5, Problem 49WE

To determine

To solve: $0=4b^2-0.48$ .

Answer to Problem 49WE

  $b=\pm 0.3$

Explanation of Solution

By using the property of square root as $x^2=\sqrt{a^2}$ implies $x=\pm a$ the given equation $0=4b^2-0.48$ can be simplified as

  $\begin{array}{l}{b}^2=\frac{0.48}{4}\\ b=\pm \sqrt{\frac{0.48}{4}}\\ b=\pm 0.3\end{array}$