How do you solve $x^{2} - 16 = 0$ $x^2-16=0$ ?

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Answer 1 · 2016-07-25T23:22:04

By using the identity $a^{2} - b^{2} = (a - b) \cdot (a + b)$ $a^2-b^2=(a-b)*(a+b)$ we have that

$x^{2} - 16 = 0 \Rightarrow x^{2} - 4^{2} = 0 \Rightarrow (x - 4) (x + 4) = 0$ $x^2-16=0=>x^2-4^2=0=>(x-4)(x+4)=0$

Hence the solutions are $x_{1} = - 4, x_{2} = 4$ $x_1=-4,x_2=4$

How do you solve x2−16=0x^2-16=0?