How do you solve $25x^2-64$ ?

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Answer 1 · 2015-06-28T08:31:48

This is a difference of squares:

$25x^2-64 = (5x)^2 - 8^2 = (5x-8)(5x+8)$

So the roots of $25x^2-64 = 0$ are $8/5$ and $-8/5$

Any difference of squares is of the form $a^2-b^2 = (a-b)(a+b)$

In our case $a = 5x$ and $b=8$ , giving:

$25x^2-64 = (5x)^2 - 8^2$

$=a^2 - b^2$

$=(a-b)(a+b)$

$= (5x-8)(5x+8)$

How do you solve 25x^2-6425x^2-64?