Explore: Comparing Solving Methods Solutions

Comparing Solving Methods Solutions

Equations written in the general form ax2+c=0 can be solved in two ways: 
- After isolating $x^{2}$ , find the square root . 
  $x^{2} = d x = \pm \sqrt{d}$

- Factor a difference of two squares.
  $\begin{matrix} x^{2} - d = (x + \sqrt{d}) (x - \sqrt{d}) \end{matrix}$  
When two identical square roots are multiplied together, the result is the radicand . 
Therefore, even if a number is not a perfect square, it can still be factored as a difference of two squares.
A difference of two squares reveals another pattern, conjugates . 
Polynomial equations can have solutions that are: real, imaginary, or complex .  
To factor equations in $a x^{2} + c = 0$ form, rewrite them as: $a x^{2} - (- c) = 0$ and factor using the difference of two squares.

Example 4

Solve and compare methods.

$x^{2} + 2 = 0$

Square Root

$\begin{array}{rcl} x^{2} & = & - 2 \\ \sqrt{x^{2}} & = & \pm \sqrt{- 2} \end{array}$

$\begin{array}{rcl} x & = & \pm i \sqrt{2} \end{array}$

Factoring

$x^{2} - (- 2) = 0 (x + \sqrt{- 2}) (x - \sqrt{- 2}) = 0 x + i \sqrt{2} = 0, x - i \sqrt{2} = 0$

$\begin{array}{rcl} x & = & \pm i \sqrt{2} \end{array}$

Explain

Both solving methods result in the same answer because $x = \pm \sqrt{d}$ and the difference of two squares are both ways to represent conjugate pairs.

Example 5

Solve by finding the square root and factoring.

$x^{2} - 2 = 0$

$\begin{array}{rcl} x^{2} & = & 2 \\ \sqrt{x^{2}} & = & \pm \sqrt{2} \end{array}$

$\begin{array}{rcl} x & = & \pm \sqrt{2} \end{array}$

$(x + \sqrt{2}) (x - \sqrt{2}) = 0 x + \sqrt{2} = 0, x - \sqrt{2} = 0$

$\begin{array}{rcl} x & = & \pm \sqrt{2} \end{array}$