How do you solve $x^{2} - 25 = 0$ $x^2 − 25 = 0$ ?

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Answer 1 · 2016-05-06T12:02:53

$x = \pm 5$ $x=+-5$

There are two ways of doing this.

$Method 1$ $color(blue)("Method 1")$

Add 25 to both sides

$x^{2} - 25 + 25 = 25$ $x^2-25+25=25$

$but - 25 + 25 = 0$ $"but "-25+25=0$

$x^{2} = 25$ $x^2=25$

$x = \sqrt{25} = \pm 5$ $color(blue)(x=sqrt(25)=+-5)$
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$Method 2$ $color(blue)("Method 2")$

Using $a^{2} - b^{2} = (a + b) (a - b)$ $a^2-b^2=(a+b)(a-b)$

Write $x^{2} - 25 = 0 as x^{2} - 5^{2} = 0 \to (x - 5) (x + 5) = 0$ $" " x^2-25=0" as "x^2-5^2 = 0 ->(x-5)(x+5)=0$

$x = \pm 5$ $color(blue)(x=+-5)$

How do you solve x2−25=0x^2 − 25 = 0?