Solve sqrt{48}-3/2sqrt{27}+sqrt{243}

$$\sqrt { 48 } - \frac { 3 } { 2 } \sqrt { 27 } + \sqrt { 243 }$$

Evaluate

$$\frac{17\sqrt{3}}{2}\approx 14.722431864$$

Solution Steps

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}.

$$4\sqrt{3}-\frac{3}{2}\sqrt{27}+\sqrt{243}$$

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}.

$$4\sqrt{3}-\frac{3}{2}\times 3\sqrt{3}+\sqrt{243}$$

Express -\frac{3}{2}\times 3 as a single fraction.

$$4\sqrt{3}+\frac{-3\times 3}{2}\sqrt{3}+\sqrt{243}$$

Multiply -3 and 3 to get -9.

$$4\sqrt{3}+\frac{-9}{2}\sqrt{3}+\sqrt{243}$$

Fraction \frac{-9}{2} can be rewritten as -\frac{9}{2} by extracting the negative sign.

$$4\sqrt{3}-\frac{9}{2}\sqrt{3}+\sqrt{243}$$

Combine 4\sqrt{3} and -\frac{9}{2}\sqrt{3} to get -\frac{1}{2}\sqrt{3}.

$$-\frac{1}{2}\sqrt{3}+\sqrt{243}$$

Factor 243=9^{2}\times 3. Rewrite the square root of the product \sqrt{9^{2}\times 3} as the product of square roots \sqrt{9^{2}}\sqrt{3}. Take the square root of 9^{2}.

$$-\frac{1}{2}\sqrt{3}+9\sqrt{3}$$

Combine -\frac{1}{2}\sqrt{3} and 9\sqrt{3} to get \frac{17}{2}\sqrt{3}.

$$\frac{17}{2}\sqrt{3}$$