Chapter 9.6, Problem 22PPS

To determine

The solutions of the equation $81x^2=9$ using quadratic formula

Answer to Problem 22PPS

  $x=\frac{1}{3},\text{ and }x=\frac{-1}{3}$

Explanation of Solution

Given:

The equation,

  $81x^2=9$

Concept Used:

Quadratic formula: The roots of a quadratic equation $a{x}^2+bx+c=0;\text{ where }a\ne 0$ are:

  $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Calculation:

In order to solve the given equation $81x^2-9=0$ , since it is in the descending order of exponents, so comparing with the general equation $a{x}^2+bx+c=0$ , the value of the coefficient a, b, and c is given by

  $a=81,b=0,\text{ and }c=-9$

Now, above quadratic equation $81y^2+0y-9=0$ can be solved using quadratic formula as shown below,

  $\begin{array}{c}x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\ =\frac{-\left(0\right)\pm \sqrt{{\left(0\right)}^2-4\left(81\right)\left(-9\right)}}{2\left(81\right)}\\ =\frac{0\pm \sqrt{0+2916}}{162}\\ =\frac{\pm \sqrt{2916}}{162}\\ =\frac{\pm 54}{162}\\ x=\frac{54}{162},\text{ and }x=\frac{-54}{162}\\ \Rightarrow x=\frac{1}{3},\text{ and }x=\frac{-1}{3}\\ \Rightarrow x=\frac{-6}{5},\text{ and }x=-3\end{array}$

Thus, the solutions of the given question are, $x=\frac{1}{3},\text{ and }x=\frac{-1}{3}$ .