Solve sqrt{12}-sqrt{63}+224/sqrt{28}

$$\sqrt{ 12 } - \sqrt{ 63 } + \frac{ 224 }{ \sqrt{ 28 } }$$

Evaluate

$$2\sqrt{3}+13\sqrt{7}\approx 37.858868659$$

Solution Steps

Factor 12=2^{2}\times 3. Rewrite the square root of the product \sqrt{2^{2}\times 3} as the product of square roots \sqrt{2^{2}}\sqrt{3}. Take the square root of 2^{2}.

$$2\sqrt{3}-\sqrt{63}+\frac{224}{\sqrt{28}}$$

Factor 63=3^{2}\times 7. Rewrite the square root of the product \sqrt{3^{2}\times 7} as the product of square roots \sqrt{3^{2}}\sqrt{7}. Take the square root of 3^{2}.

$$2\sqrt{3}-3\sqrt{7}+\frac{224}{\sqrt{28}}$$

Factor 28=2^{2}\times 7. Rewrite the square root of the product \sqrt{2^{2}\times 7} as the product of square roots \sqrt{2^{2}}\sqrt{7}. Take the square root of 2^{2}.

$$2\sqrt{3}-3\sqrt{7}+\frac{224}{2\sqrt{7}}$$

Rationalize the denominator of \frac{224}{2\sqrt{7}} by multiplying numerator and denominator by \sqrt{7}.

$$2\sqrt{3}-3\sqrt{7}+\frac{224\sqrt{7}}{2\left(\sqrt{7}\right)^{2}}$$

The square of \sqrt{7} is 7.

$$2\sqrt{3}-3\sqrt{7}+\frac{224\sqrt{7}}{2\times 7}$$

Cancel out 2\times 7 in both numerator and denominator.

$$2\sqrt{3}-3\sqrt{7}+16\sqrt{7}$$

Combine -3\sqrt{7} and 16\sqrt{7} to get 13\sqrt{7}.

$$2\sqrt{3}+13\sqrt{7}$$