How do you solve $x^{2} - 75 = 0$ $x^2-75=0$ ?

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How do you solve $x^{2} - 75 = 0$ $x^2-75=0$ ?

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Answer 1 · 2016-08-17T21:51:33

$x = \pm 5 \sqrt{3}$ $x=+-5sqrt(3)$

Explanation:

Treat $x^{2} - 75$ $x^2-75$ as the difference of squares

We know that $a^{2} - b^{2} = (a - b) (a + b)$ $a^2-b^2=(a-b)(a+b)$

So replacing $a$ $a$ with $x$ $x$
and $b$ $b$ with $5 \sqrt{3}$ $5sqrt(3)$ (so that $b^{2} = 75$ $b^2=75$ )
we have
$XXX x^{2} - 75 = 0$ $color(white)("XXX")x^2-75=0$
is equivalent to
$XXX (x - 5 \sqrt{3}) (x + 5 \sqrt{3}) = 0$ $color(white)("XXX")(x-5sqrt(3))(x+5sqrt(3))=0$

which implies
either $(x - 5 \sqrt{3}) = 0 XX or XX (x + 5 \sqrt{3}) = 0$ $(x-5sqrt(3))=0 color(white)("XX")orcolor(white)("XX") (x+5sqrt(3))=0$
$XX \to x = 5 \sqrt{3} XXXXXXXX \to x = - 5 \sqrt{3}$ $color(white)("XX")rarr x=5sqrt(3)color(white)("XXXXXXXX")rarrx=-5sqrt(3)$

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How do you solve $x^{2} - 75 = 0$ $x^2-75=0$ ?

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How do you solve $x^{2} - 75 = 0$ $x^2-75=0$ ?