How do you solve $x^{4} - 81 = 0$ $x^4-81=0$ ?

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Answer 1 · 2015-09-29T12:25:11

$x = \pm 3$ $color(blue)(x=+-3$

$x^{4} - 81 = 0$ $x^4 -81=0$

By property
$a^{2} - b^{2} = (a - b) (a + b)$ $color(blue)(a^2-b^2=(a-b)(a+b)$

Similarly we can rewrite the expression given :

$x^{4} - 81 = {(x^{2})}^{2} - (9^{2})$ $x^4 -81=(x^2)^2-(9^2)$

$= (x^{2} - 9) (x^{2} + 9)$ $=color(blue)((x^2-9)(x^2+9)$

Equating the factors to zero:

1) $x^{2} - 9 = 0$ $x^2-9=0$
$x^{2} = 9$ $x^2=9$
$x = \pm 3$ $color(blue)(x=+-3$

2) $x^{2} + 9 = 0$ $x^2+9=0$
$x^{2} = - 9$ $x^2=-9$
This case is not applicable as we cannot take the root of a negative number.

How do you solve x4−81=0x^4-81=0?