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

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

Evaluate

$$\sqrt{7}+2\sqrt{3}\approx 6.109852926$$

Solution Steps

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}.

$$3\sqrt{7}+\sqrt{12}-\sqrt{28}$$

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}.

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

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}.

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

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

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