Solve sqrt{50}+sqrt{8}=sqrt{98}

$$\sqrt { 50 } + \sqrt { 8 } = \sqrt { 98 }$$

Verify

Solution Steps

Factor 50=5^{2}\times 2. Rewrite the square root of the product \sqrt{5^{2}\times 2} as the product of square roots \sqrt{5^{2}}\sqrt{2}. Take the square root of 5^{2}.

$$5\sqrt{2}+\sqrt{8}=\sqrt{98}$$

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

$$5\sqrt{2}+2\sqrt{2}=\sqrt{98}$$

Combine 5\sqrt{2} and 2\sqrt{2} to get 7\sqrt{2}.

$$7\sqrt{2}=\sqrt{98}$$

Factor 98=7^{2}\times 2. Rewrite the square root of the product \sqrt{7^{2}\times 2} as the product of square roots \sqrt{7^{2}}\sqrt{2}. Take the square root of 7^{2}.

$$7\sqrt{2}=7\sqrt{2}$$

Subtract 7\sqrt{2} from both sides.

$$7\sqrt{2}-7\sqrt{2}=0$$

Combine 7\sqrt{2} and -7\sqrt{2} to get 0.

$$0=0$$

Compare 0 and 0.

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