Chapter 9.6, Problem 17PPS

To determine

The solutions of the equation $x^2+16=0$ using quadratic formula

Answer to Problem 17PPS

  $x=\pm 2,\text{ and }x=\pm 2$

Explanation of Solution

Given:

The equation,

  $x^2+16=0$

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 $x^2+16=0$ , 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

  $a=1,b=0,\text{ and }c=16$

Now, above quadratic equation $1y^2+0\cdot y+16=0$ can be solved using quadratic formula as shown below,

  $\begin{array}{c}x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\ =\frac{-\left(0\right)\pm \sqrt{{\left(0\right)}^2-4\left(1\right)\left(16\right)}}{2\left(1\right)}\\ =\frac{0\pm \sqrt{0-16}}{2}\\ =\frac{0\pm \sqrt{-16}}{8}\\ =\frac{\pm 4}{2}\\ =\pm 2\\ =2,\text{ and }y=-2\\ \Rightarrow y=2,\text{ and }y=-2\end{array}$

Thus, the solutions of the given question are, $x=\pm 2,\text{ and }x=\pm 2$ .