Solve 1(sqrt{243}+sqrt{128})

$$1 ( \sqrt { 243 } + \sqrt { 128 } )$$

Evaluate

$$8\sqrt{2}+9\sqrt{3}\approx 26.902165767$$

Solution Steps

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}.

$$1\left(9\sqrt{3}+\sqrt{128}\right)$$

Factor 128=8^{2}\times 2. Rewrite the square root of the product \sqrt{8^{2}\times 2} as the product of square roots \sqrt{8^{2}}\sqrt{2}. Take the square root of 8^{2}.

$$1\left(9\sqrt{3}+8\sqrt{2}\right)$$

Use the distributive property to multiply 1 by 9\sqrt{3}+8\sqrt{2}.

$$9\sqrt{3}+8\sqrt{2}$$