Solve 25x^2=4?

$$25 x ^ { 2 } = 4 ?$$

Solve for x

$$x=\frac{2}{5}=0.4$$

$$x=-\frac{2}{5}=-0.4$$

Steps Using Difference of Squares

Divide both sides by 25.

$$x^{2}=\frac{4}{25}$$

Subtract \frac{4}{25} from both sides.

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

Multiply both sides by 25.

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

Consider 25x^{2}-4. Rewrite 25x^{2}-4 as \left(5x\right)^{2}-2^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

$$\left(5x-2\right)\left(5x+2\right)=0$$

To find equation solutions, solve 5x-2=0 and 5x+2=0.

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

Steps by Finding Square Root

Divide both sides by 25.

$$x^{2}=\frac{4}{25}$$

Take the square root of both sides of the equation.

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

Steps Using the Quadratic Formula

Divide both sides by 25.

$$x^{2}=\frac{4}{25}$$

Subtract \frac{4}{25} from both sides.

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

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 0 for b, and -\frac{4}{25} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{0±\sqrt{0^{2}-4\left(-\frac{4}{25}\right)}}{2}$$

Square 0.

$$x=\frac{0±\sqrt{-4\left(-\frac{4}{25}\right)}}{2}$$

Multiply -4 times -\frac{4}{25}.

$$x=\frac{0±\sqrt{\frac{16}{25}}}{2}$$

Take the square root of \frac{16}{25}.

$$x=\frac{0±\frac{4}{5}}{2}$$

Now solve the equation x=\frac{0±\frac{4}{5}}{2} when ± is plus.

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

Now solve the equation x=\frac{0±\frac{4}{5}}{2} when ± is minus.

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

The equation is now solved.

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