Solve (3sqrt{3}-2sqrt{5})^2=

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

Evaluate

$$47-12\sqrt{15}\approx 0.524199846$$

Solution Steps

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

$$9\left(\sqrt{3}\right)^{2}-12\sqrt{3}\sqrt{5}+4\left(\sqrt{5}\right)^{2}$$

The square of \sqrt{3} is 3.

$$9\times 3-12\sqrt{3}\sqrt{5}+4\left(\sqrt{5}\right)^{2}$$

Multiply 9 and 3 to get 27.

$$27-12\sqrt{3}\sqrt{5}+4\left(\sqrt{5}\right)^{2}$$

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

$$27-12\sqrt{15}+4\left(\sqrt{5}\right)^{2}$$

The square of \sqrt{5} is 5.

$$27-12\sqrt{15}+4\times 5$$

Multiply 4 and 5 to get 20.

$$27-12\sqrt{15}+20$$

Add 27 and 20 to get 47.

$$47-12\sqrt{15}$$

Expand

$$47-12\sqrt{15}$$

Solution Steps

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

$$9\left(\sqrt{3}\right)^{2}-12\sqrt{3}\sqrt{5}+4\left(\sqrt{5}\right)^{2}$$

The square of \sqrt{3} is 3.

$$9\times 3-12\sqrt{3}\sqrt{5}+4\left(\sqrt{5}\right)^{2}$$

Multiply 9 and 3 to get 27.

$$27-12\sqrt{3}\sqrt{5}+4\left(\sqrt{5}\right)^{2}$$

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

$$27-12\sqrt{15}+4\left(\sqrt{5}\right)^{2}$$

The square of \sqrt{5} is 5.

$$27-12\sqrt{15}+4\times 5$$

Multiply 4 and 5 to get 20.

$$27-12\sqrt{15}+20$$

Add 27 and 20 to get 47.

$$47-12\sqrt{15}$$