Solve 7:(2+sqrt{3})^2-(2-sqrt{3})^2

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

Evaluate

$$42-24\sqrt{3}\approx 0.430780618$$

Solution Steps

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

$$\frac{7}{4+4\sqrt{3}+\left(\sqrt{3}\right)^{2}}-\left(2-\sqrt{3}\right)^{2}$$

The square of \sqrt{3} is 3.

$$\frac{7}{4+4\sqrt{3}+3}-\left(2-\sqrt{3}\right)^{2}$$

Add 4 and 3 to get 7.

$$\frac{7}{7+4\sqrt{3}}-\left(2-\sqrt{3}\right)^{2}$$

Rationalize the denominator of \frac{7}{7+4\sqrt{3}} by multiplying numerator and denominator by 7-4\sqrt{3}.

$$\frac{7\left(7-4\sqrt{3}\right)}{\left(7+4\sqrt{3}\right)\left(7-4\sqrt{3}\right)}-\left(2-\sqrt{3}\right)^{2}$$

Consider \left(7+4\sqrt{3}\right)\left(7-4\sqrt{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}.

$$\frac{7\left(7-4\sqrt{3}\right)}{7^{2}-\left(4\sqrt{3}\right)^{2}}-\left(2-\sqrt{3}\right)^{2}$$

Calculate 7 to the power of 2 and get 49.

$$\frac{7\left(7-4\sqrt{3}\right)}{49-\left(4\sqrt{3}\right)^{2}}-\left(2-\sqrt{3}\right)^{2}$$

Expand \left(4\sqrt{3}\right)^{2}.

$$\frac{7\left(7-4\sqrt{3}\right)}{49-4^{2}\left(\sqrt{3}\right)^{2}}-\left(2-\sqrt{3}\right)^{2}$$

Calculate 4 to the power of 2 and get 16.

$$\frac{7\left(7-4\sqrt{3}\right)}{49-16\left(\sqrt{3}\right)^{2}}-\left(2-\sqrt{3}\right)^{2}$$

The square of \sqrt{3} is 3.

$$\frac{7\left(7-4\sqrt{3}\right)}{49-16\times 3}-\left(2-\sqrt{3}\right)^{2}$$

Multiply 16 and 3 to get 48.

$$\frac{7\left(7-4\sqrt{3}\right)}{49-48}-\left(2-\sqrt{3}\right)^{2}$$

Subtract 48 from 49 to get 1.

$$\frac{7\left(7-4\sqrt{3}\right)}{1}-\left(2-\sqrt{3}\right)^{2}$$

Anything divided by one gives itself.

$$7\left(7-4\sqrt{3}\right)-\left(2-\sqrt{3}\right)^{2}$$

Use the distributive property to multiply 7 by 7-4\sqrt{3}.

$$49-28\sqrt{3}-\left(2-\sqrt{3}\right)^{2}$$

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

$$49-28\sqrt{3}-\left(4-4\sqrt{3}+\left(\sqrt{3}\right)^{2}\right)$$

The square of \sqrt{3} is 3.

$$49-28\sqrt{3}-\left(4-4\sqrt{3}+3\right)$$

Add 4 and 3 to get 7.

$$49-28\sqrt{3}-\left(7-4\sqrt{3}\right)$$

To find the opposite of 7-4\sqrt{3}, find the opposite of each term.

$$49-28\sqrt{3}-7+4\sqrt{3}$$

Subtract 7 from 49 to get 42.

$$42-28\sqrt{3}+4\sqrt{3}$$

Combine -28\sqrt{3} and 4\sqrt{3} to get -24\sqrt{3}.

$$42-24\sqrt{3}$$

Expand

$$42-24\sqrt{3}$$

Solution Steps

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

$$\frac{7}{4+4\sqrt{3}+\left(\sqrt{3}\right)^{2}}-\left(2-\sqrt{3}\right)^{2}$$

The square of \sqrt{3} is 3.

$$\frac{7}{4+4\sqrt{3}+3}-\left(2-\sqrt{3}\right)^{2}$$

Add 4 and 3 to get 7.

$$\frac{7}{7+4\sqrt{3}}-\left(2-\sqrt{3}\right)^{2}$$

Rationalize the denominator of \frac{7}{7+4\sqrt{3}} by multiplying numerator and denominator by 7-4\sqrt{3}.

$$\frac{7\left(7-4\sqrt{3}\right)}{\left(7+4\sqrt{3}\right)\left(7-4\sqrt{3}\right)}-\left(2-\sqrt{3}\right)^{2}$$

Consider \left(7+4\sqrt{3}\right)\left(7-4\sqrt{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}.

$$\frac{7\left(7-4\sqrt{3}\right)}{7^{2}-\left(4\sqrt{3}\right)^{2}}-\left(2-\sqrt{3}\right)^{2}$$

Calculate 7 to the power of 2 and get 49.

$$\frac{7\left(7-4\sqrt{3}\right)}{49-\left(4\sqrt{3}\right)^{2}}-\left(2-\sqrt{3}\right)^{2}$$

Expand \left(4\sqrt{3}\right)^{2}.

$$\frac{7\left(7-4\sqrt{3}\right)}{49-4^{2}\left(\sqrt{3}\right)^{2}}-\left(2-\sqrt{3}\right)^{2}$$

Calculate 4 to the power of 2 and get 16.

$$\frac{7\left(7-4\sqrt{3}\right)}{49-16\left(\sqrt{3}\right)^{2}}-\left(2-\sqrt{3}\right)^{2}$$

The square of \sqrt{3} is 3.

$$\frac{7\left(7-4\sqrt{3}\right)}{49-16\times 3}-\left(2-\sqrt{3}\right)^{2}$$

Multiply 16 and 3 to get 48.

$$\frac{7\left(7-4\sqrt{3}\right)}{49-48}-\left(2-\sqrt{3}\right)^{2}$$

Subtract 48 from 49 to get 1.

$$\frac{7\left(7-4\sqrt{3}\right)}{1}-\left(2-\sqrt{3}\right)^{2}$$

Anything divided by one gives itself.

$$7\left(7-4\sqrt{3}\right)-\left(2-\sqrt{3}\right)^{2}$$

Use the distributive property to multiply 7 by 7-4\sqrt{3}.

$$49-28\sqrt{3}-\left(2-\sqrt{3}\right)^{2}$$

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

$$49-28\sqrt{3}-\left(4-4\sqrt{3}+\left(\sqrt{3}\right)^{2}\right)$$

The square of \sqrt{3} is 3.

$$49-28\sqrt{3}-\left(4-4\sqrt{3}+3\right)$$

Add 4 and 3 to get 7.

$$49-28\sqrt{3}-\left(7-4\sqrt{3}\right)$$

To find the opposite of 7-4\sqrt{3}, find the opposite of each term.

$$49-28\sqrt{3}-7+4\sqrt{3}$$

Subtract 7 from 49 to get 42.

$$42-28\sqrt{3}+4\sqrt{3}$$

Combine -28\sqrt{3} and 4\sqrt{3} to get -24\sqrt{3}.

$$42-24\sqrt{3}$$