Solve (sqrt{18}+sqrt{50}):sqrt{2}+(sqrt{5}+1)^2+(sqrt{5}-1)(sqrt{5}+1)

$$( \sqrt { 18 } + \sqrt { 50 } ) : \sqrt { 2 } + ( \sqrt { 5 } + 1 ) ^ { 2 } + ( \sqrt { 5 } - 1 ) ( \sqrt { 5 } + 1 )$$

Evaluate

$$2\left(\sqrt{5}+9\right)\approx 22.472135955$$

Solution Steps

Factor 18=3^{2}\times 2. Rewrite the square root of the product \sqrt{3^{2}\times 2} as the product of square roots \sqrt{3^{2}}\sqrt{2}. Take the square root of 3^{2}.

$$\frac{3\sqrt{2}+\sqrt{50}}{\sqrt{2}}+\left(\sqrt{5}+1\right)^{2}+\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)$$

Factor 50=5^{2}\times 2. Rewrite the square root of the product \sqrt{5^{2}\times 2} as the product of square roots \sqrt{5^{2}}\sqrt{2}. Take the square root of 5^{2}.

$$\frac{3\sqrt{2}+5\sqrt{2}}{\sqrt{2}}+\left(\sqrt{5}+1\right)^{2}+\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)$$

Combine 3\sqrt{2} and 5\sqrt{2} to get 8\sqrt{2}.

$$\frac{8\sqrt{2}}{\sqrt{2}}+\left(\sqrt{5}+1\right)^{2}+\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)$$

Rationalize the denominator of \frac{8\sqrt{2}}{\sqrt{2}} by multiplying numerator and denominator by \sqrt{2}.

$$\frac{8\sqrt{2}\sqrt{2}}{\left(\sqrt{2}\right)^{2}}+\left(\sqrt{5}+1\right)^{2}+\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)$$

The square of \sqrt{2} is 2.

$$\frac{8\sqrt{2}\sqrt{2}}{2}+\left(\sqrt{5}+1\right)^{2}+\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)$$

Multiply \sqrt{2} and \sqrt{2} to get 2.

$$\frac{8\times 2}{2}+\left(\sqrt{5}+1\right)^{2}+\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)$$

Multiply 8 and 2 to get 16.

$$\frac{16}{2}+\left(\sqrt{5}+1\right)^{2}+\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)$$

Divide 16 by 2 to get 8.

$$8+\left(\sqrt{5}+1\right)^{2}+\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(\sqrt{5}+1\right)^{2}.

$$8+\left(\sqrt{5}\right)^{2}+2\sqrt{5}+1+\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)$$

The square of \sqrt{5} is 5.

$$8+5+2\sqrt{5}+1+\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)$$

Add 5 and 1 to get 6.

$$8+6+2\sqrt{5}+\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)$$

Add 8 and 6 to get 14.

$$14+2\sqrt{5}+\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)$$

Consider \left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 1.

$$14+2\sqrt{5}+\left(\sqrt{5}\right)^{2}-1$$

The square of \sqrt{5} is 5.

$$14+2\sqrt{5}+5-1$$

Subtract 1 from 5 to get 4.

$$14+2\sqrt{5}+4$$

Add 14 and 4 to get 18.

$$18+2\sqrt{5}$$