How do you solve $9x^2 - 16 = 0$ ?

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Answer 1 · 2016-04-20T17:11:57

$(3x+4)(3x-4)$

In the general form $(a+b)(a-b)=a^2-b^2$

Treating $9x^2$ as $a^2$
$rArr a=3x$
and
treating $16$ as $b^2$
$rArr b=4$

$color(green)((3x+4)(3x-4))=9x^2-16$

How do you solve 9x^2 - 16 = 09x^2 - 16 = 0?