Solve {left(3sqrt{5}-2sqrt{15}right)}^2

$${ \left(3 \sqrt{ 5 } -2 \sqrt{ 15 } \right) }^{ 2 }$$

Evaluate

$$105-60\sqrt{3}\approx 1.076951546$$

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(3\sqrt{5}-2\sqrt{15}\right)^{2}.

$$9\left(\sqrt{5}\right)^{2}-12\sqrt{5}\sqrt{15}+4\left(\sqrt{15}\right)^{2}$$

The square of \sqrt{5} is 5.

$$9\times 5-12\sqrt{5}\sqrt{15}+4\left(\sqrt{15}\right)^{2}$$

Multiply 9 and 5 to get 45.

$$45-12\sqrt{5}\sqrt{15}+4\left(\sqrt{15}\right)^{2}$$

Factor 15=5\times 3. Rewrite the square root of the product \sqrt{5\times 3} as the product of square roots \sqrt{5}\sqrt{3}.

$$45-12\sqrt{5}\sqrt{5}\sqrt{3}+4\left(\sqrt{15}\right)^{2}$$

Multiply \sqrt{5} and \sqrt{5} to get 5.

$$45-12\times 5\sqrt{3}+4\left(\sqrt{15}\right)^{2}$$

Multiply -12 and 5 to get -60.

$$45-60\sqrt{3}+4\left(\sqrt{15}\right)^{2}$$

The square of \sqrt{15} is 15.

$$45-60\sqrt{3}+4\times 15$$

Multiply 4 and 15 to get 60.

$$45-60\sqrt{3}+60$$

Add 45 and 60 to get 105.

$$105-60\sqrt{3}$$

Expand

$$105-60\sqrt{3}$$

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(3\sqrt{5}-2\sqrt{15}\right)^{2}.

$$9\left(\sqrt{5}\right)^{2}-12\sqrt{5}\sqrt{15}+4\left(\sqrt{15}\right)^{2}$$

The square of \sqrt{5} is 5.

$$9\times 5-12\sqrt{5}\sqrt{15}+4\left(\sqrt{15}\right)^{2}$$

Multiply 9 and 5 to get 45.

$$45-12\sqrt{5}\sqrt{15}+4\left(\sqrt{15}\right)^{2}$$

Factor 15=5\times 3. Rewrite the square root of the product \sqrt{5\times 3} as the product of square roots \sqrt{5}\sqrt{3}.

$$45-12\sqrt{5}\sqrt{5}\sqrt{3}+4\left(\sqrt{15}\right)^{2}$$

Multiply \sqrt{5} and \sqrt{5} to get 5.

$$45-12\times 5\sqrt{3}+4\left(\sqrt{15}\right)^{2}$$

Multiply -12 and 5 to get -60.

$$45-60\sqrt{3}+4\left(\sqrt{15}\right)^{2}$$

The square of \sqrt{15} is 15.

$$45-60\sqrt{3}+4\times 15$$

Multiply 4 and 15 to get 60.

$$45-60\sqrt{3}+60$$

Add 45 and 60 to get 105.

$$105-60\sqrt{3}$$