Solve (sqrt{2}+sqrt{3})^2+(4-sqrt{24})^2

$$( \sqrt { 2 } + \sqrt { 3 } ) ^ { 2 } + ( 4 - \sqrt { 24 } ) ^ { 2 }$$

Evaluate

$$45-14\sqrt{6}\approx 10.707143601$$

Solution Steps

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

$$\left(\sqrt{2}\right)^{2}+2\sqrt{2}\sqrt{3}+\left(\sqrt{3}\right)^{2}+\left(4-\sqrt{24}\right)^{2}$$

The square of \sqrt{2} is 2.

$$2+2\sqrt{2}\sqrt{3}+\left(\sqrt{3}\right)^{2}+\left(4-\sqrt{24}\right)^{2}$$

To multiply \sqrt{2} and \sqrt{3}, multiply the numbers under the square root.

$$2+2\sqrt{6}+\left(\sqrt{3}\right)^{2}+\left(4-\sqrt{24}\right)^{2}$$

The square of \sqrt{3} is 3.

$$2+2\sqrt{6}+3+\left(4-\sqrt{24}\right)^{2}$$

Add 2 and 3 to get 5.

$$5+2\sqrt{6}+\left(4-\sqrt{24}\right)^{2}$$

Factor 24=2^{2}\times 6. Rewrite the square root of the product \sqrt{2^{2}\times 6} as the product of square roots \sqrt{2^{2}}\sqrt{6}. Take the square root of 2^{2}.

$$5+2\sqrt{6}+\left(4-2\sqrt{6}\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(4-2\sqrt{6}\right)^{2}.

$$5+2\sqrt{6}+16-16\sqrt{6}+4\left(\sqrt{6}\right)^{2}$$

The square of \sqrt{6} is 6.

$$5+2\sqrt{6}+16-16\sqrt{6}+4\times 6$$

Multiply 4 and 6 to get 24.

$$5+2\sqrt{6}+16-16\sqrt{6}+24$$

Add 16 and 24 to get 40.

$$5+2\sqrt{6}+40-16\sqrt{6}$$

Add 5 and 40 to get 45.

$$45+2\sqrt{6}-16\sqrt{6}$$

Combine 2\sqrt{6} and -16\sqrt{6} to get -14\sqrt{6}.

$$45-14\sqrt{6}$$

Expand

$$45-14\sqrt{6}$$

Solution Steps

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

$$\left(\sqrt{2}\right)^{2}+2\sqrt{2}\sqrt{3}+\left(\sqrt{3}\right)^{2}+\left(4-\sqrt{24}\right)^{2}$$

The square of \sqrt{2} is 2.

$$2+2\sqrt{2}\sqrt{3}+\left(\sqrt{3}\right)^{2}+\left(4-\sqrt{24}\right)^{2}$$

To multiply \sqrt{2} and \sqrt{3}, multiply the numbers under the square root.

$$2+2\sqrt{6}+\left(\sqrt{3}\right)^{2}+\left(4-\sqrt{24}\right)^{2}$$

The square of \sqrt{3} is 3.

$$2+2\sqrt{6}+3+\left(4-\sqrt{24}\right)^{2}$$

Add 2 and 3 to get 5.

$$5+2\sqrt{6}+\left(4-\sqrt{24}\right)^{2}$$

Factor 24=2^{2}\times 6. Rewrite the square root of the product \sqrt{2^{2}\times 6} as the product of square roots \sqrt{2^{2}}\sqrt{6}. Take the square root of 2^{2}.

$$5+2\sqrt{6}+\left(4-2\sqrt{6}\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(4-2\sqrt{6}\right)^{2}.

$$5+2\sqrt{6}+16-16\sqrt{6}+4\left(\sqrt{6}\right)^{2}$$

The square of \sqrt{6} is 6.

$$5+2\sqrt{6}+16-16\sqrt{6}+4\times 6$$

Multiply 4 and 6 to get 24.

$$5+2\sqrt{6}+16-16\sqrt{6}+24$$

Add 16 and 24 to get 40.

$$5+2\sqrt{6}+40-16\sqrt{6}$$

Add 5 and 40 to get 45.

$$45+2\sqrt{6}-16\sqrt{6}$$

Combine 2\sqrt{6} and -16\sqrt{6} to get -14\sqrt{6}.

$$45-14\sqrt{6}$$