Solve 9f^2-4=0

$$9 f ^ { 2 } - 4 = 0$$

Solve for f

$$f=\frac{2}{3}\approx 0.666666667$$

$$f=-\frac{2}{3}\approx -0.666666667$$

Steps Using Difference of Squares

Consider 9f^{2}-4. Rewrite 9f^{2}-4 as \left(3f\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).

$$\left(3f-2\right)\left(3f+2\right)=0$$

To find equation solutions, solve 3f-2=0 and 3f+2=0.

$$f=\frac{2}{3} f=-\frac{2}{3}$$

Steps by Finding Square Root

Add 4 to both sides. Anything plus zero gives itself.

$$9f^{2}=4$$

Divide both sides by 9.

$$f^{2}=\frac{4}{9}$$

Take the square root of both sides of the equation.

$$f=\frac{2}{3} f=-\frac{2}{3}$$

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.

$$9f^{2}-4=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 9 for a, 0 for b, and -4 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$f=\frac{0±\sqrt{0^{2}-4\times 9\left(-4\right)}}{2\times 9}$$

Square 0.

$$f=\frac{0±\sqrt{-4\times 9\left(-4\right)}}{2\times 9}$$

Multiply -4 times 9.

$$f=\frac{0±\sqrt{-36\left(-4\right)}}{2\times 9}$$

Multiply -36 times -4.

$$f=\frac{0±\sqrt{144}}{2\times 9}$$

Take the square root of 144.

$$f=\frac{0±12}{2\times 9}$$

Multiply 2 times 9.

$$f=\frac{0±12}{18}$$

Now solve the equation f=\frac{0±12}{18} when ± is plus. Reduce the fraction \frac{12}{18} to lowest terms by extracting and canceling out 6.

$$f=\frac{2}{3}$$

Now solve the equation f=\frac{0±12}{18} when ± is minus. Reduce the fraction \frac{-12}{18} to lowest terms by extracting and canceling out 6.

$$f=-\frac{2}{3}$$

The equation is now solved.

$$f=\frac{2}{3} f=-\frac{2}{3}$$