Solve (2/sqrt{10+3})^2

$$( \frac { 2 } { \sqrt { 10 } + 3 } ) ^ { 2 }$$

Evaluate

$$76-24\sqrt{10}\approx 0.105336156$$

Solution Steps

Rationalize the denominator of \frac{2}{\sqrt{10}+3} by multiplying numerator and denominator by \sqrt{10}-3.

$$\left(\frac{2\left(\sqrt{10}-3\right)}{\left(\sqrt{10}+3\right)\left(\sqrt{10}-3\right)}\right)^{2}$$

Consider \left(\sqrt{10}+3\right)\left(\sqrt{10}-3\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$\left(\frac{2\left(\sqrt{10}-3\right)}{\left(\sqrt{10}\right)^{2}-3^{2}}\right)^{2}$$

Square \sqrt{10}. Square 3.

$$\left(\frac{2\left(\sqrt{10}-3\right)}{10-9}\right)^{2}$$

Subtract 9 from 10 to get 1.

$$\left(\frac{2\left(\sqrt{10}-3\right)}{1}\right)^{2}$$

Anything divided by one gives itself.

$$\left(2\left(\sqrt{10}-3\right)\right)^{2}$$

Use the distributive property to multiply 2 by \sqrt{10}-3.

$$\left(2\sqrt{10}-6\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2\sqrt{10}-6\right)^{2}.

$$4\left(\sqrt{10}\right)^{2}-24\sqrt{10}+36$$

The square of \sqrt{10} is 10.

$$4\times 10-24\sqrt{10}+36$$

Multiply 4 and 10 to get 40.

$$40-24\sqrt{10}+36$$

Add 40 and 36 to get 76.

$$76-24\sqrt{10}$$

Expand

$$76-24\sqrt{10}$$

Solution Steps

Rationalize the denominator of \frac{2}{\sqrt{10}+3} by multiplying numerator and denominator by \sqrt{10}-3.

$$\left(\frac{2\left(\sqrt{10}-3\right)}{\left(\sqrt{10}+3\right)\left(\sqrt{10}-3\right)}\right)^{2}$$

Consider \left(\sqrt{10}+3\right)\left(\sqrt{10}-3\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$\left(\frac{2\left(\sqrt{10}-3\right)}{\left(\sqrt{10}\right)^{2}-3^{2}}\right)^{2}$$

Square \sqrt{10}. Square 3.

$$\left(\frac{2\left(\sqrt{10}-3\right)}{10-9}\right)^{2}$$

Subtract 9 from 10 to get 1.

$$\left(\frac{2\left(\sqrt{10}-3\right)}{1}\right)^{2}$$

Anything divided by one gives itself.

$$\left(2\left(\sqrt{10}-3\right)\right)^{2}$$

Use the distributive property to multiply 2 by \sqrt{10}-3.

$$\left(2\sqrt{10}-6\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2\sqrt{10}-6\right)^{2}.

$$4\left(\sqrt{10}\right)^{2}-24\sqrt{10}+36$$

The square of \sqrt{10} is 10.

$$4\times 10-24\sqrt{10}+36$$

Multiply 4 and 10 to get 40.

$$40-24\sqrt{10}+36$$

Add 40 and 36 to get 76.

$$76-24\sqrt{10}$$