Precalculus Examples

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

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

Step 1

Simplify each term.

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Step 1.1

Rewrite

28

$28$ as

2^{2} \cdot 7

$2^{2} \cdot 7$ .

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Step 1.1.1

Factor

4

$4$ out of

28

$28$ .

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

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

Step 1.1.2

Rewrite

4

$4$ as

2^{2}

$2^{2}$ .

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

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

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

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

Step 1.2

Pull terms out from under the radical.

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

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

Step 1.3

Multiply

2

$2$ by

- 1

$- 1$ .

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

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

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

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

Step 2

Since

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

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

Always true