Solve 4c^2-1=0

$$4 c ^ { 2 } - 1 = 0$$

Solve for c

$$c=\frac{1}{2}=0.5$$

$$c=-\frac{1}{2}=-0.5$$

Steps Using Difference of Squares

Consider 4c^{2}-1. Rewrite 4c^{2}-1 as \left(2c\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).

$$\left(2c-1\right)\left(2c+1\right)=0$$

To find equation solutions, solve 2c-1=0 and 2c+1=0.

$$c=\frac{1}{2} c=-\frac{1}{2}$$

Steps by Finding Square Root

Add 1 to both sides. Anything plus zero gives itself.

$$4c^{2}=1$$

Divide both sides by 4.

$$c^{2}=\frac{1}{4}$$

Take the square root of both sides of the equation.

$$c=\frac{1}{2} c=-\frac{1}{2}$$

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.

$$4c^{2}-1=0$$

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

$$c=\frac{0±\sqrt{0^{2}-4\times 4\left(-1\right)}}{2\times 4}$$

Square 0.

$$c=\frac{0±\sqrt{-4\times 4\left(-1\right)}}{2\times 4}$$

Multiply -4 times 4.

$$c=\frac{0±\sqrt{-16\left(-1\right)}}{2\times 4}$$

Multiply -16 times -1.

$$c=\frac{0±\sqrt{16}}{2\times 4}$$

Take the square root of 16.

$$c=\frac{0±4}{2\times 4}$$

Multiply 2 times 4.

$$c=\frac{0±4}{8}$$

Now solve the equation c=\frac{0±4}{8} when ± is plus. Reduce the fraction \frac{4}{8} to lowest terms by extracting and canceling out 4.

$$c=\frac{1}{2}$$

Now solve the equation c=\frac{0±4}{8} when ± is minus. Reduce the fraction \frac{-4}{8} to lowest terms by extracting and canceling out 4.

$$c=-\frac{1}{2}$$

The equation is now solved.

$$c=\frac{1}{2} c=-\frac{1}{2}$$