Solve sqrt{75}+sqrt{63}-sqrt{48}-14/sqrt{7}

$$\sqrt{ 75 } + \sqrt{ 63 } - \sqrt{ 48 } - \frac{ 14 }{ \sqrt{ 7 } }$$

Evaluate

$$\sqrt{3}+\sqrt{7}\approx 4.377802119$$

Solution Steps

Factor 75=5^{2}\times 3. Rewrite the square root of the product \sqrt{5^{2}\times 3} as the product of square roots \sqrt{5^{2}}\sqrt{3}. Take the square root of 5^{2}.

$$5\sqrt{3}+\sqrt{63}-\sqrt{48}-\frac{14}{\sqrt{7}}$$

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}.

$$5\sqrt{3}+3\sqrt{7}-\sqrt{48}-\frac{14}{\sqrt{7}}$$

Factor 48=4^{2}\times 3. Rewrite the square root of the product \sqrt{4^{2}\times 3} as the product of square roots \sqrt{4^{2}}\sqrt{3}. Take the square root of 4^{2}.

$$5\sqrt{3}+3\sqrt{7}-4\sqrt{3}-\frac{14}{\sqrt{7}}$$

Combine 5\sqrt{3} and -4\sqrt{3} to get \sqrt{3}.

$$\sqrt{3}+3\sqrt{7}-\frac{14}{\sqrt{7}}$$

Rationalize the denominator of \frac{14}{\sqrt{7}} by multiplying numerator and denominator by \sqrt{7}.

$$\sqrt{3}+3\sqrt{7}-\frac{14\sqrt{7}}{\left(\sqrt{7}\right)^{2}}$$

The square of \sqrt{7} is 7.

$$\sqrt{3}+3\sqrt{7}-\frac{14\sqrt{7}}{7}$$

Divide 14\sqrt{7} by 7 to get 2\sqrt{7}.

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

Combine 3\sqrt{7} and -2\sqrt{7} to get \sqrt{7}.

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