Solve (frac{3+2sqrt{3}}{2})^2+(frac{10-3sqrt{3}}{2})^2

$$( \frac { 3 + 2 \sqrt { 3 } } { 2 } ) ^ { 2 } + ( \frac { 10 - 3 \sqrt { 3 } } { 2 } ) ^ { 2 }$$

Evaluate

$$37-12\sqrt{3}\approx 16.215390309$$

Short Solution Steps

To raise \frac{3+2\sqrt{3}}{2} to a power, raise both numerator and denominator to the power and then divide.

$$\frac{\left(3+2\sqrt{3}\right)^{2}}{2^{2}}+\left(\frac{10-3\sqrt{3}}{2}\right)^{2}$$

To raise \frac{10-3\sqrt{3}}{2} to a power, raise both numerator and denominator to the power and then divide.

$$\frac{\left(3+2\sqrt{3}\right)^{2}}{2^{2}}+\frac{\left(10-3\sqrt{3}\right)^{2}}{2^{2}}$$

Since \frac{\left(3+2\sqrt{3}\right)^{2}}{2^{2}} and \frac{\left(10-3\sqrt{3}\right)^{2}}{2^{2}} have the same denominator, add them by adding their numerators.

$$\frac{\left(3+2\sqrt{3}\right)^{2}+\left(10-3\sqrt{3}\right)^{2}}{2^{2}}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(3+2\sqrt{3}\right)^{2}.

$$\frac{9+12\sqrt{3}+4\left(\sqrt{3}\right)^{2}}{2^{2}}+\frac{\left(10-3\sqrt{3}\right)^{2}}{2^{2}}$$

The square of \sqrt{3} is 3.

$$\frac{9+12\sqrt{3}+4\times 3}{2^{2}}+\frac{\left(10-3\sqrt{3}\right)^{2}}{2^{2}}$$

Multiply 4 and 3 to get 12.

$$\frac{9+12\sqrt{3}+12}{2^{2}}+\frac{\left(10-3\sqrt{3}\right)^{2}}{2^{2}}$$

Add 9 and 12 to get 21.

$$\frac{21+12\sqrt{3}}{2^{2}}+\frac{\left(10-3\sqrt{3}\right)^{2}}{2^{2}}$$

Calculate 2 to the power of 2 and get 4.

$$\frac{21+12\sqrt{3}}{4}+\frac{\left(10-3\sqrt{3}\right)^{2}}{2^{2}}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(10-3\sqrt{3}\right)^{2}.

$$\frac{21+12\sqrt{3}}{4}+\frac{100-60\sqrt{3}+9\left(\sqrt{3}\right)^{2}}{2^{2}}$$

The square of \sqrt{3} is 3.

$$\frac{21+12\sqrt{3}}{4}+\frac{100-60\sqrt{3}+9\times 3}{2^{2}}$$

Multiply 9 and 3 to get 27.

$$\frac{21+12\sqrt{3}}{4}+\frac{100-60\sqrt{3}+27}{2^{2}}$$

Add 100 and 27 to get 127.

$$\frac{21+12\sqrt{3}}{4}+\frac{127-60\sqrt{3}}{2^{2}}$$

Calculate 2 to the power of 2 and get 4.

$$\frac{21+12\sqrt{3}}{4}+\frac{127-60\sqrt{3}}{4}$$

Since \frac{21+12\sqrt{3}}{4} and \frac{127-60\sqrt{3}}{4} have the same denominator, add them by adding their numerators.

$$\frac{21+12\sqrt{3}+127-60\sqrt{3}}{4}$$

Do the calculations in 21+12\sqrt{3}+127-60\sqrt{3}.

$$\frac{148-48\sqrt{3}}{4}$$

Divide each term of 148-48\sqrt{3} by 4 to get 37-12\sqrt{3}.

$$37-12\sqrt{3}$$

Expand

$$37-12\sqrt{3}$$

Short Solution Steps

To raise \frac{3+2\sqrt{3}}{2} to a power, raise both numerator and denominator to the power and then divide.

$$\frac{\left(3+2\sqrt{3}\right)^{2}}{2^{2}}+\left(\frac{10-3\sqrt{3}}{2}\right)^{2}$$

To raise \frac{10-3\sqrt{3}}{2} to a power, raise both numerator and denominator to the power and then divide.

$$\frac{\left(3+2\sqrt{3}\right)^{2}}{2^{2}}+\frac{\left(10-3\sqrt{3}\right)^{2}}{2^{2}}$$

Since \frac{\left(3+2\sqrt{3}\right)^{2}}{2^{2}} and \frac{\left(10-3\sqrt{3}\right)^{2}}{2^{2}} have the same denominator, add them by adding their numerators.

$$\frac{\left(3+2\sqrt{3}\right)^{2}+\left(10-3\sqrt{3}\right)^{2}}{2^{2}}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(3+2\sqrt{3}\right)^{2}.

$$\frac{9+12\sqrt{3}+4\left(\sqrt{3}\right)^{2}}{2^{2}}+\frac{\left(10-3\sqrt{3}\right)^{2}}{2^{2}}$$

The square of \sqrt{3} is 3.

$$\frac{9+12\sqrt{3}+4\times 3}{2^{2}}+\frac{\left(10-3\sqrt{3}\right)^{2}}{2^{2}}$$

Multiply 4 and 3 to get 12.

$$\frac{9+12\sqrt{3}+12}{2^{2}}+\frac{\left(10-3\sqrt{3}\right)^{2}}{2^{2}}$$

Add 9 and 12 to get 21.

$$\frac{21+12\sqrt{3}}{2^{2}}+\frac{\left(10-3\sqrt{3}\right)^{2}}{2^{2}}$$

Calculate 2 to the power of 2 and get 4.

$$\frac{21+12\sqrt{3}}{4}+\frac{\left(10-3\sqrt{3}\right)^{2}}{2^{2}}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(10-3\sqrt{3}\right)^{2}.

$$\frac{21+12\sqrt{3}}{4}+\frac{100-60\sqrt{3}+9\left(\sqrt{3}\right)^{2}}{2^{2}}$$

The square of \sqrt{3} is 3.

$$\frac{21+12\sqrt{3}}{4}+\frac{100-60\sqrt{3}+9\times 3}{2^{2}}$$

Multiply 9 and 3 to get 27.

$$\frac{21+12\sqrt{3}}{4}+\frac{100-60\sqrt{3}+27}{2^{2}}$$

Add 100 and 27 to get 127.

$$\frac{21+12\sqrt{3}}{4}+\frac{127-60\sqrt{3}}{2^{2}}$$

Calculate 2 to the power of 2 and get 4.

$$\frac{21+12\sqrt{3}}{4}+\frac{127-60\sqrt{3}}{4}$$

Since \frac{21+12\sqrt{3}}{4} and \frac{127-60\sqrt{3}}{4} have the same denominator, add them by adding their numerators.

$$\frac{21+12\sqrt{3}+127-60\sqrt{3}}{4}$$

Do the calculations in 21+12\sqrt{3}+127-60\sqrt{3}.

$$\frac{148-48\sqrt{3}}{4}$$

Divide each term of 148-48\sqrt{3} by 4 to get 37-12\sqrt{3}.

$$37-12\sqrt{3}$$