Solve (sqrt{15}+sqrt{3})^2

$$( \sqrt { 15 } + \sqrt { 3 } ) ^ { 2 }$$

Evaluate

$$6\left(\sqrt{5}+3\right)\approx 31.416407865$$

Solution Steps

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

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

The square of \sqrt{15} is 15.

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

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

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

Multiply \sqrt{3} and \sqrt{3} to get 3.

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

Multiply 2 and 3 to get 6.

$$15+6\sqrt{5}+\left(\sqrt{3}\right)^{2}$$

The square of \sqrt{3} is 3.

$$15+6\sqrt{5}+3$$

Add 15 and 3 to get 18.

$$18+6\sqrt{5}$$

Expand

$$6 \sqrt{5} + 18 = 31.416407865$$

Solution Steps

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

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

The square of \sqrt{15} is 15.

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

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

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

Multiply \sqrt{3} and \sqrt{3} to get 3.

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

Multiply 2 and 3 to get 6.

$$15+6\sqrt{5}+\left(\sqrt{3}\right)^{2}$$

The square of \sqrt{3} is 3.

$$15+6\sqrt{5}+3$$

Add 15 and 3 to get 18.

$$18+6\sqrt{5}$$