Solve t^2+(-3/4t^2-9/4t)^2

$$t ^ { 2 } + ( - \frac { 3 } { 4 } t ^ { 2 } - \frac { 9 } { 4 } t ) ^ { 2 }$$

Evaluate

$$\frac{t^{2}\left(9t^{2}+54t+97\right)}{16}$$

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(-\frac{3}{4}t^{2}-\frac{9}{4}t\right)^{2}.

$$t^{2}+\frac{9}{16}\left(t^{2}\right)^{2}+\frac{27}{8}t^{2}t+\frac{81}{16}t^{2}$$

To raise a power to another power, multiply the exponents. Multiply 2 and 2 to get 4.

$$t^{2}+\frac{9}{16}t^{4}+\frac{27}{8}t^{2}t+\frac{81}{16}t^{2}$$

To multiply powers of the same base, add their exponents. Add 2 and 1 to get 3.

$$t^{2}+\frac{9}{16}t^{4}+\frac{27}{8}t^{3}+\frac{81}{16}t^{2}$$

Combine t^{2} and \frac{81}{16}t^{2} to get \frac{97}{16}t^{2}.

$$\frac{97}{16}t^{2}+\frac{9}{16}t^{4}+\frac{27}{8}t^{3}$$

Expand

$$\frac{9t^{4}}{16}+\frac{27t^{3}}{8}+\frac{97t^{2}}{16}$$

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(-\frac{3}{4}t^{2}-\frac{9}{4}t\right)^{2}.

$$t^{2}+\frac{9}{16}\left(t^{2}\right)^{2}+\frac{27}{8}t^{2}t+\frac{81}{16}t^{2}$$

To raise a power to another power, multiply the exponents. Multiply 2 and 2 to get 4.

$$t^{2}+\frac{9}{16}t^{4}+\frac{27}{8}t^{2}t+\frac{81}{16}t^{2}$$

To multiply powers of the same base, add their exponents. Add 2 and 1 to get 3.

$$t^{2}+\frac{9}{16}t^{4}+\frac{27}{8}t^{3}+\frac{81}{16}t^{2}$$

Combine t^{2} and \frac{81}{16}t^{2} to get \frac{97}{16}t^{2}.

$$\frac{97}{16}t^{2}+\frac{9}{16}t^{4}+\frac{27}{8}t^{3}$$