Solve (sqrt{6}+sqrt{3})(sqrt{6}+sqrt{3})

$$( \sqrt { 6 } + \sqrt { 3 } ) ( \sqrt { 6 } + \sqrt { 3 } )$$

Evaluate

$$3\left(2\sqrt{2}+3\right)\approx 17.485281374$$

Solution Steps

Multiply \sqrt{6}+\sqrt{3} and \sqrt{6}+\sqrt{3} to get \left(\sqrt{6}+\sqrt{3}\right)^{2}.

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

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

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

The square of \sqrt{6} is 6.

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

Factor 6=3\times 2. Rewrite the square root of the product \sqrt{3\times 2} as the product of square roots \sqrt{3}\sqrt{2}.

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

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

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

Multiply 2 and 3 to get 6.

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

The square of \sqrt{3} is 3.

$$6+6\sqrt{2}+3$$

Add 6 and 3 to get 9.

$$9+6\sqrt{2}$$

Factor

$$3 {(2 \sqrt{2} + 3)} = 17.485281374$$