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Solve each of the quadratic equations below. Show your work. x2 - 16 = 0
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\[a^2 - b^2 = (a+b)(a-b)\] are you familiar with this?
^ To continue: \(\Large \color{purple}{\rightarrow (x + 4)(x - 4) = 0 }\) We can say, that either (x + 4) is 0, or (x - 4) is 0, as a number multiplied by 0 is 0. So: 1) x + 4 = 0 x = -4 or 2) x - 4 = 0 x = 4 x can be either 4 or -4 with this.
How did I get four? \(\Large \color{purple}{\rightarrow a^2 = 16 }\) \(\Large \color{purple}{\rightarrow a = \sqrt{16} = 4 }\)
\[X ^{2}-16=0\] \[X ^{2}=16\] \[X=\pm4\]
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