Solve frac{sqrt{24}+sqrt{216}}{sqrt{96}}=3

$$\frac { \sqrt { 24 } + \sqrt { 216 } } { \sqrt { 96 } } = 3$$

Verify

Solution Steps

Factor 24=2^{2}\times 6. Rewrite the square root of the product \sqrt{2^{2}\times 6} as the product of square roots \sqrt{2^{2}}\sqrt{6}. Take the square root of 2^{2}.

$$\frac{2\sqrt{6}+\sqrt{216}}{\sqrt{96}}=3$$

Factor 216=6^{2}\times 6. Rewrite the square root of the product \sqrt{6^{2}\times 6} as the product of square roots \sqrt{6^{2}}\sqrt{6}. Take the square root of 6^{2}.

$$\frac{2\sqrt{6}+6\sqrt{6}}{\sqrt{96}}=3$$

Combine 2\sqrt{6} and 6\sqrt{6} to get 8\sqrt{6}.

$$\frac{8\sqrt{6}}{\sqrt{96}}=3$$

Factor 96=4^{2}\times 6. Rewrite the square root of the product \sqrt{4^{2}\times 6} as the product of square roots \sqrt{4^{2}}\sqrt{6}. Take the square root of 4^{2}.

$$\frac{8\sqrt{6}}{4\sqrt{6}}=3$$

Cancel out 4\sqrt{6} in both numerator and denominator.

$$2=3$$

Compare 2 and 3.

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