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# How do you factor and solve 25x^2 -9 =0?

Jul 12, 2016

$x \in \left\{- \frac{3}{5} , \frac{3}{5}\right\}$

#### Explanation:

Using the difference of squares formula ${a}^{2} - {b}^{2} = \left(a + b\right) \left(a - b\right)$, we have

$25 {x}^{2} - 9 = {\left(5 x\right)}^{2} - {3}^{2} = \left(5 x + 3\right) \left(5 x - 3\right) = 0$

$\implies 5 x + 3 = 0 \mathmr{and} 5 x - 3 = 0$

$\implies 5 x = - 3 \mathmr{and} 5 x = 3$

$\implies x = - \frac{3}{5} \mathmr{and} x = \frac{3}{5}$

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