Solve by Factoring 25x^2-4=0

$25 x^{2} - 4 = 0$

Step 1

Rewrite $25 x^{2}$ as ${(5 x)}^{2}$ .

${(5 x)}^{2} - 4 = 0$

Step 2

Rewrite $4$ as $2^{2}$ .

${(5 x)}^{2} - 2^{2} = 0$

Step 3

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 5 x$ and $b = 2$ .

$(5 x + 2) (5 x - 2) = 0$

Step 4

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$5 x + 2 = 0$

$5 x - 2 = 0$

Step 5

Set $5 x + 2$ equal to $0$ and solve for $x$ .

Set $5 x + 2$ equal to $0$ .

$5 x + 2 = 0$

Solve $5 x + 2 = 0$ for $x$ .

Subtract $2$ from both sides of the equation.

$5 x = - 2$

Divide each term in $5 x = - 2$ by $5$ and simplify.

Divide each term in $5 x = - 2$ by $5$ .

$\frac{5 x}{5} = \frac{- 2}{5}$

Simplify the left side.

Cancel the common factor of $5$ .

Cancel the common factor.

$\frac{5 x}{5} = \frac{- 2}{5}$

Divide $x$ by $1$ .

$x = \frac{- 2}{5}$

Simplify the right side.

Move the negative in front of the fraction.

$x = - \frac{2}{5}$

Step 6

Set $5 x - 2$ equal to $0$ and solve for $x$ .

Set $5 x - 2$ equal to $0$ .

$5 x - 2 = 0$

Solve $5 x - 2 = 0$ for $x$ .

Add $2$ to both sides of the equation.

$5 x = 2$

Divide each term in $5 x = 2$ by $5$ and simplify.

Divide each term in $5 x = 2$ by $5$ .

$\frac{5 x}{5} = \frac{2}{5}$

Simplify the left side.

Cancel the common factor of $5$ .

Cancel the common factor.

$\frac{5 x}{5} = \frac{2}{5}$

Divide $x$ by $1$ .

$x = \frac{2}{5}$

Step 7

The final solution is all the values that make $(5 x + 2) (5 x - 2) = 0$ true.

$x = - \frac{2}{5}, \frac{2}{5}$

Solve by Factoring 25x^2-4=0

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