Precalculus Examples

${(\sqrt{7} - 1)}^{2} = 8 - \sqrt{28}$

Step 1

Simplify each term.

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Step 1.1

Rewrite $28$ as $2^{2} \cdot 7$ .

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Step 1.1.1

Factor $4$ out of $28$ .

${(\sqrt{7} - 1)}^{2} = 8 - \sqrt{4 (7)}$

Step 1.1.2

Rewrite $4$ as $2^{2}$ .

${(\sqrt{7} - 1)}^{2} = 8 - \sqrt{2^{2} \cdot 7}$

Step 1.2

Pull terms out from under the radical.

${(\sqrt{7} - 1)}^{2} = 8 - (2 \sqrt{7})$

Step 1.3

Multiply $2$ by $- 1$ .

${(\sqrt{7} - 1)}^{2} = 8 - 2 \sqrt{7}$

Step 2

Since ${(\sqrt{7} - 1)}^{2} = 8 - 2 \sqrt{7}$ , the equation will always be true.

Always true