Solve t^2leq(t/2)^2*3+1

$$t ^ { 2 } \leq ( \frac { t } { 2 } ) ^ { 2 } \times 3 + 1$$

Solve for t

$$t\in \begin{bmatrix}-2,2\end{bmatrix}$$

Solution Steps

To raise \frac{t}{2} to a power, raise both numerator and denominator to the power and then divide.

$$t^{2}\leq \frac{t^{2}}{2^{2}}\times 3+1$$

Express \frac{t^{2}}{2^{2}}\times 3 as a single fraction.

$$t^{2}\leq \frac{t^{2}\times 3}{2^{2}}+1$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 1 times \frac{2^{2}}{2^{2}}.

$$t^{2}\leq \frac{t^{2}\times 3}{2^{2}}+\frac{2^{2}}{2^{2}}$$

Since \frac{t^{2}\times 3}{2^{2}} and \frac{2^{2}}{2^{2}} have the same denominator, add them by adding their numerators.

$$t^{2}\leq \frac{t^{2}\times 3+2^{2}}{2^{2}}$$

Combine like terms in t^{2}\times 3+2^{2}.

$$t^{2}\leq \frac{3t^{2}+4}{2^{2}}$$

Calculate 2 to the power of 2 and get 4.

$$t^{2}\leq \frac{3t^{2}+4}{4}$$

Divide each term of 3t^{2}+4 by 4 to get \frac{3}{4}t^{2}+1.

$$t^{2}\leq \frac{3}{4}t^{2}+1$$

Subtract \frac{3}{4}t^{2} from both sides.

$$t^{2}-\frac{3}{4}t^{2}\leq 1$$

Combine t^{2} and -\frac{3}{4}t^{2} to get \frac{1}{4}t^{2}.

$$\frac{1}{4}t^{2}\leq 1$$

Multiply both sides by 4, the reciprocal of \frac{1}{4}. Since \frac{1}{4} is positive, the inequality direction remains the same.

$$t^{2}\leq 1\times 4$$

Multiply 1 and 4 to get 4.

$$t^{2}\leq 4$$

Calculate the square root of 4 and get 2. Rewrite 4 as 2^{2}.

$$t^{2}\leq 2^{2}$$

Inequality holds for |t|\leq 2.

$$|t|\leq 2$$

Rewrite |t|\leq 2 as t\in \left[-2,2\right].

$$t\in \begin{bmatrix}-2,2\end{bmatrix}$$