Solve 4r^2-81=0

$$4 r ^ { 2 } - 81 = 0$$

Solve for r

$$r = \frac{9}{2} = 4\frac{1}{2} = 4.5$$

$$r = -\frac{9}{2} = -4\frac{1}{2} = -4.5$$

Steps Using Difference of Squares

Consider 4r^{2}-81. Rewrite 4r^{2}-81 as \left(2r\right)^{2}-9^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

$$\left(2r-9\right)\left(2r+9\right)=0$$

To find equation solutions, solve 2r-9=0 and 2r+9=0.

$$r=\frac{9}{2} r=-\frac{9}{2}$$

Steps by Finding Square Root

Add 81 to both sides. Anything plus zero gives itself.

$$4r^{2}=81$$

Divide both sides by 4.

$$r^{2}=\frac{81}{4}$$

Take the square root of both sides of the equation.

$$r=\frac{9}{2} r=-\frac{9}{2}$$

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.

$$4r^{2}-81=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, 0 for b, and -81 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$r=\frac{0±\sqrt{0^{2}-4\times 4\left(-81\right)}}{2\times 4}$$

Square 0.

$$r=\frac{0±\sqrt{-4\times 4\left(-81\right)}}{2\times 4}$$

Multiply -4 times 4.

$$r=\frac{0±\sqrt{-16\left(-81\right)}}{2\times 4}$$

Multiply -16 times -81.

$$r=\frac{0±\sqrt{1296}}{2\times 4}$$

Take the square root of 1296.

$$r=\frac{0±36}{2\times 4}$$

Multiply 2 times 4.

$$r=\frac{0±36}{8}$$

Now solve the equation r=\frac{0±36}{8} when ± is plus. Reduce the fraction \frac{36}{8} to lowest terms by extracting and canceling out 4.

$$r=\frac{9}{2}$$

Now solve the equation r=\frac{0±36}{8} when ± is minus. Reduce the fraction \frac{-36}{8} to lowest terms by extracting and canceling out 4.

$$r=-\frac{9}{2}$$

The equation is now solved.

$$r=\frac{9}{2} r=-\frac{9}{2}$$