Solve (3sqrt{6}-sqrt{15})^2

$$( 3 \sqrt { 6 } - \sqrt { 15 } ) ^ { 2 }$$

Evaluate

$$69-18\sqrt{10}\approx 12.079002117$$

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(3\sqrt{6}-\sqrt{15}\right)^{2}.

$$9\left(\sqrt{6}\right)^{2}-6\sqrt{6}\sqrt{15}+\left(\sqrt{15}\right)^{2}$$

The square of \sqrt{6} is 6.

$$9\times 6-6\sqrt{6}\sqrt{15}+\left(\sqrt{15}\right)^{2}$$

Multiply 9 and 6 to get 54.

$$54-6\sqrt{6}\sqrt{15}+\left(\sqrt{15}\right)^{2}$$

To multiply \sqrt{6} and \sqrt{15}, multiply the numbers under the square root.

$$54-6\sqrt{90}+\left(\sqrt{15}\right)^{2}$$

The square of \sqrt{15} is 15.

$$54-6\sqrt{90}+15$$

Add 54 and 15 to get 69.

$$69-6\sqrt{90}$$

Factor 90=3^{2}\times 10. Rewrite the square root of the product \sqrt{3^{2}\times 10} as the product of square roots \sqrt{3^{2}}\sqrt{10}. Take the square root of 3^{2}.

$$69-6\times 3\sqrt{10}$$

Multiply -6 and 3 to get -18.

$$69-18\sqrt{10}$$

Expand

$$69-18\sqrt{10}$$

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(3\sqrt{6}-\sqrt{15}\right)^{2}.

$$9\left(\sqrt{6}\right)^{2}-6\sqrt{6}\sqrt{15}+\left(\sqrt{15}\right)^{2}$$

The square of \sqrt{6} is 6.

$$9\times 6-6\sqrt{6}\sqrt{15}+\left(\sqrt{15}\right)^{2}$$

Multiply 9 and 6 to get 54.

$$54-6\sqrt{6}\sqrt{15}+\left(\sqrt{15}\right)^{2}$$

To multiply \sqrt{6} and \sqrt{15}, multiply the numbers under the square root.

$$54-6\sqrt{90}+\left(\sqrt{15}\right)^{2}$$

The square of \sqrt{15} is 15.

$$54-6\sqrt{90}+15$$

Add 54 and 15 to get 69.

$$69-6\sqrt{90}$$

Factor 90=3^{2}\times 10. Rewrite the square root of the product \sqrt{3^{2}\times 10} as the product of square roots \sqrt{3^{2}}\sqrt{10}. Take the square root of 3^{2}.

$$69-6\times 3\sqrt{10}$$

Multiply -6 and 3 to get -18.

$$69-18\sqrt{10}$$