Solve {left(3sqrt{5}-5sqrt{3}right)}^2

$${ \left(3 \sqrt{ 5 } -5 \sqrt{ 3 } \right) }^{ 2 }$$

Evaluate

$$120-30\sqrt{15}\approx 3.810499614$$

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(3\sqrt{5}-5\sqrt{3}\right)^{2}.

$$9\left(\sqrt{5}\right)^{2}-30\sqrt{5}\sqrt{3}+25\left(\sqrt{3}\right)^{2}$$

The square of \sqrt{5} is 5.

$$9\times 5-30\sqrt{5}\sqrt{3}+25\left(\sqrt{3}\right)^{2}$$

Multiply 9 and 5 to get 45.

$$45-30\sqrt{5}\sqrt{3}+25\left(\sqrt{3}\right)^{2}$$

To multiply \sqrt{5} and \sqrt{3}, multiply the numbers under the square root.

$$45-30\sqrt{15}+25\left(\sqrt{3}\right)^{2}$$

The square of \sqrt{3} is 3.

$$45-30\sqrt{15}+25\times 3$$

Multiply 25 and 3 to get 75.

$$45-30\sqrt{15}+75$$

Add 45 and 75 to get 120.

$$120-30\sqrt{15}$$

Expand

$$120-30\sqrt{15}$$

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(3\sqrt{5}-5\sqrt{3}\right)^{2}.

$$9\left(\sqrt{5}\right)^{2}-30\sqrt{5}\sqrt{3}+25\left(\sqrt{3}\right)^{2}$$

The square of \sqrt{5} is 5.

$$9\times 5-30\sqrt{5}\sqrt{3}+25\left(\sqrt{3}\right)^{2}$$

Multiply 9 and 5 to get 45.

$$45-30\sqrt{5}\sqrt{3}+25\left(\sqrt{3}\right)^{2}$$

To multiply \sqrt{5} and \sqrt{3}, multiply the numbers under the square root.

$$45-30\sqrt{15}+25\left(\sqrt{3}\right)^{2}$$

The square of \sqrt{3} is 3.

$$45-30\sqrt{15}+25\times 3$$

Multiply 25 and 3 to get 75.

$$45-30\sqrt{15}+75$$

Add 45 and 75 to get 120.

$$120-30\sqrt{15}$$