Solve T^2/24^2=1/4^2

$$\frac { T ^ { 2 } } { 24 ^ { 2 } } = \frac { 1 } { 4 ^ { 2 } }$$

Solve for T

$$T=6$$

$$T=-6$$

Steps Using Difference of Squares

Calculate 24 to the power of 2 and get 576.

$$\frac{T^{2}}{576}=\frac{1}{4^{2}}$$

Calculate 4 to the power of 2 and get 16.

$$\frac{T^{2}}{576}=\frac{1}{16}$$

Subtract \frac{1}{16} from both sides.

$$\frac{T^{2}}{576}-\frac{1}{16}=0$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 576 and 16 is 576. Multiply \frac{1}{16} times \frac{36}{36}.

$$\frac{T^{2}}{576}-\frac{36}{576}=0$$

Since \frac{T^{2}}{576} and \frac{36}{576} have the same denominator, subtract them by subtracting their numerators.

$$\frac{T^{2}-36}{576}=0$$

Divide each term of T^{2}-36 by 576 to get \frac{1}{576}T^{2}-\frac{1}{16}.

$$\frac{1}{576}T^{2}-\frac{1}{16}=0$$

Multiply both sides by 576.

$$T^{2}-36=0$$

Consider T^{2}-36. Rewrite T^{2}-36 as T^{2}-6^{2}. The difference of squares can be factored using the rule: a^{2}-b^{2}=\left(a-b\right)\left(a+b\right).

$$\left(T-6\right)\left(T+6\right)=0$$

To find equation solutions, solve T-6=0 and T+6=0.

$$T=6 T=-6$$

Steps by Finding Square Root

Calculate 24 to the power of 2 and get 576.

$$\frac{T^{2}}{576}=\frac{1}{4^{2}}$$

Calculate 4 to the power of 2 and get 16.

$$\frac{T^{2}}{576}=\frac{1}{16}$$

Multiply both sides by 576.

$$T^{2}=\frac{1}{16}\times 576$$

Multiply \frac{1}{16} and 576 to get 36.

$$T^{2}=36$$

Take the square root of both sides of the equation.

$$T=6 T=-6$$

Steps Using the Quadratic Formula

Calculate 24 to the power of 2 and get 576.

$$\frac{T^{2}}{576}=\frac{1}{4^{2}}$$

Calculate 4 to the power of 2 and get 16.

$$\frac{T^{2}}{576}=\frac{1}{16}$$

Subtract \frac{1}{16} from both sides.

$$\frac{T^{2}}{576}-\frac{1}{16}=0$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 576 and 16 is 576. Multiply \frac{1}{16} times \frac{36}{36}.

$$\frac{T^{2}}{576}-\frac{36}{576}=0$$

Since \frac{T^{2}}{576} and \frac{36}{576} have the same denominator, subtract them by subtracting their numerators.

$$\frac{T^{2}-36}{576}=0$$

Divide each term of T^{2}-36 by 576 to get \frac{1}{576}T^{2}-\frac{1}{16}.

$$\frac{1}{576}T^{2}-\frac{1}{16}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute \frac{1}{576} for a, 0 for b, and -\frac{1}{16} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$T=\frac{0±\sqrt{0^{2}-4\times \frac{1}{576}\left(-\frac{1}{16}\right)}}{2\times \frac{1}{576}}$$

Square 0.

$$T=\frac{0±\sqrt{-4\times \frac{1}{576}\left(-\frac{1}{16}\right)}}{2\times \frac{1}{576}}$$

Multiply -4 times \frac{1}{576}.

$$T=\frac{0±\sqrt{-\frac{1}{144}\left(-\frac{1}{16}\right)}}{2\times \frac{1}{576}}$$

Multiply -\frac{1}{144} times -\frac{1}{16} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$T=\frac{0±\sqrt{\frac{1}{2304}}}{2\times \frac{1}{576}}$$

Take the square root of \frac{1}{2304}.

$$T=\frac{0±\frac{1}{48}}{2\times \frac{1}{576}}$$

Multiply 2 times \frac{1}{576}.

$$T=\frac{0±\frac{1}{48}}{\frac{1}{288}}$$

Now solve the equation T=\frac{0±\frac{1}{48}}{\frac{1}{288}} when ± is plus.

$$T=6$$

Now solve the equation T=\frac{0±\frac{1}{48}}{\frac{1}{288}} when ± is minus.

$$T=-6$$

The equation is now solved.

$$T=6 T=-6$$