Solve (7sqrt{5}-5sqrt{3})^2

$$( 7 \sqrt { 5 } - 5 \sqrt { 3 } ) ^ { 2 }$$

Evaluate

$$320-70\sqrt{15}\approx 48.891165765$$

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(7\sqrt{5}-5\sqrt{3}\right)^{2}.

$$49\left(\sqrt{5}\right)^{2}-70\sqrt{5}\sqrt{3}+25\left(\sqrt{3}\right)^{2}$$

The square of \sqrt{5} is 5.

$$49\times 5-70\sqrt{5}\sqrt{3}+25\left(\sqrt{3}\right)^{2}$$

Multiply 49 and 5 to get 245.

$$245-70\sqrt{5}\sqrt{3}+25\left(\sqrt{3}\right)^{2}$$

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

$$245-70\sqrt{15}+25\left(\sqrt{3}\right)^{2}$$

The square of \sqrt{3} is 3.

$$245-70\sqrt{15}+25\times 3$$

Multiply 25 and 3 to get 75.

$$245-70\sqrt{15}+75$$

Add 245 and 75 to get 320.

$$320-70\sqrt{15}$$

Expand

$$320-70\sqrt{15}$$

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(7\sqrt{5}-5\sqrt{3}\right)^{2}.

$$49\left(\sqrt{5}\right)^{2}-70\sqrt{5}\sqrt{3}+25\left(\sqrt{3}\right)^{2}$$

The square of \sqrt{5} is 5.

$$49\times 5-70\sqrt{5}\sqrt{3}+25\left(\sqrt{3}\right)^{2}$$

Multiply 49 and 5 to get 245.

$$245-70\sqrt{5}\sqrt{3}+25\left(\sqrt{3}\right)^{2}$$

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

$$245-70\sqrt{15}+25\left(\sqrt{3}\right)^{2}$$

The square of \sqrt{3} is 3.

$$245-70\sqrt{15}+25\times 3$$

Multiply 25 and 3 to get 75.

$$245-70\sqrt{15}+75$$

Add 245 and 75 to get 320.

$$320-70\sqrt{15}$$