Solve sqrt{27}+(sqrt{3}-2)^2+9^frac{1{2}}-(pi-3.14)^0

$$\sqrt { 27 } + ( \sqrt { 3 } - 2 ) ^ { 2 } + 9 ^ { \frac { 1 } { 2 } } - ( \pi - 3.14 ) ^ { 0 }$$

Evaluate

$$9-\sqrt{3}\approx 7.267949192$$

Solution Steps

Factor 27=3^{2}\times 3. Rewrite the square root of the product \sqrt{3^{2}\times 3} as the product of square roots \sqrt{3^{2}}\sqrt{3}. Take the square root of 3^{2}.

$$3\sqrt{3}+\left(\sqrt{3}-2\right)^{2}+9^{\frac{1}{2}}-\left(\pi -3.14\right)^{0}$$

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

$$3\sqrt{3}+\left(\sqrt{3}\right)^{2}-4\sqrt{3}+4+9^{\frac{1}{2}}-\left(\pi -3.14\right)^{0}$$

The square of \sqrt{3} is 3.

$$3\sqrt{3}+3-4\sqrt{3}+4+9^{\frac{1}{2}}-\left(\pi -3.14\right)^{0}$$

Add 3 and 4 to get 7.

$$3\sqrt{3}+7-4\sqrt{3}+9^{\frac{1}{2}}-\left(\pi -3.14\right)^{0}$$

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

$$-\sqrt{3}+7+9^{\frac{1}{2}}-\left(\pi -3.14\right)^{0}$$

Calculate 9 to the power of \frac{1}{2} and get 3.

$$-\sqrt{3}+7+3-\left(\pi -3.14\right)^{0}$$

Add 7 and 3 to get 10.

$$-\sqrt{3}+10-\left(\pi -3.14\right)^{0}$$

Calculate \pi -3.14 to the power of 0 and get 1.

$$-\sqrt{3}+10-1$$

Subtract 1 from 10 to get 9.

$$-\sqrt{3}+9$$