Solve 125x^2-5=0

$$125 x ^ { 2 } - 5 = 0$$

Solve for x

$$x=\frac{1}{5}=0.2$$

$$x=-\frac{1}{5}=-0.2$$

Steps Using Difference of Squares

Divide both sides by 5.

$$25x^{2}-1=0$$

Consider 25x^{2}-1. Rewrite 25x^{2}-1 as \left(5x\right)^{2}-1^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

$$\left(5x-1\right)\left(5x+1\right)=0$$

To find equation solutions, solve 5x-1=0 and 5x+1=0.

$$x=\frac{1}{5} x=-\frac{1}{5}$$

Steps by Finding Square Root

Add 5 to both sides. Anything plus zero gives itself.

$$125x^{2}=5$$

Divide both sides by 125.

$$x^{2}=\frac{5}{125}$$

Reduce the fraction \frac{5}{125} to lowest terms by extracting and canceling out 5.

$$x^{2}=\frac{1}{25}$$

Take the square root of both sides of the equation.

$$x=\frac{1}{5} x=-\frac{1}{5}$$

Steps Using the Quadratic Formula

Quadratic equations like this one, with an x^{2} term but no x term, can still be solved using the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}, once they are put in standard form: ax^{2}+bx+c=0.

$$125x^{2}-5=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 125 for a, 0 for b, and -5 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{0±\sqrt{0^{2}-4\times 125\left(-5\right)}}{2\times 125}$$

Square 0.

$$x=\frac{0±\sqrt{-4\times 125\left(-5\right)}}{2\times 125}$$

Multiply -4 times 125.

$$x=\frac{0±\sqrt{-500\left(-5\right)}}{2\times 125}$$

Multiply -500 times -5.

$$x=\frac{0±\sqrt{2500}}{2\times 125}$$

Take the square root of 2500.

$$x=\frac{0±50}{2\times 125}$$

Multiply 2 times 125.

$$x=\frac{0±50}{250}$$

Now solve the equation x=\frac{0±50}{250} when ± is plus. Reduce the fraction \frac{50}{250} to lowest terms by extracting and canceling out 50.

$$x=\frac{1}{5}$$

Now solve the equation x=\frac{0±50}{250} when ± is minus. Reduce the fraction \frac{-50}{250} to lowest terms by extracting and canceling out 50.

$$x=-\frac{1}{5}$$

The equation is now solved.

$$x=\frac{1}{5} x=-\frac{1}{5}$$