How do you solve a^4-1=0?

Jan 6, 2016

$a = 1 , - 1$

Explanation:

${a}^{4} - 1 = 0$

${a}^{4} = 1$

The only solutions of a could be 1 and -1 because,

$1 \cdot 1 \cdot 1 \cdot 1 = 1$

$- 1 \cdot - 1 \cdot - 1 \cdot - 1 = 1$

Jan 6, 2016

$a = \pm 1$

Explanation:

alternatively

$\left({a}^{4} - 1\right) = \left({a}^{2} + 1\right) \times \left({a}^{2} - 1\right) = \left({a}^{2} + 1\right) \times \left(a + 1\right) \times \left(a - 1\right) = 0$
Now a moltiplation is zero when the factors are zero
$\left({a}^{2} + 1\right) = 0$ $\cancel{\exists} x \in \mathbb{R}$ (no Real solutions)
$\left(a + 1\right) = 0$; $a = - 1$
$\left(a - 1\right) = 0$; $a = + 1$