Simplify (7- square root of 2)(8+ square root of 2)

$(7 - \sqrt{2}) (8 + \sqrt{2})$

Step 1

Expand $(7 - \sqrt{2}) (8 + \sqrt{2})$ using the FOIL Method.

Apply the distributive property.

$7 (8 + \sqrt{2}) - \sqrt{2} (8 + \sqrt{2})$

Apply the distributive property.

$7 \cdot 8 + 7 \sqrt{2} - \sqrt{2} (8 + \sqrt{2})$

Apply the distributive property.

$7 \cdot 8 + 7 \sqrt{2} - \sqrt{2} \cdot 8 - \sqrt{2} \sqrt{2}$

Step 2

Simplify and combine like terms.

Simplify each term.

Multiply $7$ by $8$ .

$56 + 7 \sqrt{2} - \sqrt{2} \cdot 8 - \sqrt{2} \sqrt{2}$

Multiply $8$ by $- 1$ .

$56 + 7 \sqrt{2} - 8 \sqrt{2} - \sqrt{2} \sqrt{2}$

Multiply $- \sqrt{2} \sqrt{2}$ .

Raise $\sqrt{2}$ to the power of $1$ .

$56 + 7 \sqrt{2} - 8 \sqrt{2} - ({\sqrt{2}}^{1} \sqrt{2})$

Raise $\sqrt{2}$ to the power of $1$ .

$56 + 7 \sqrt{2} - 8 \sqrt{2} - ({\sqrt{2}}^{1} {\sqrt{2}}^{1})$

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$56 + 7 \sqrt{2} - 8 \sqrt{2} - {\sqrt{2}}^{1 + 1}$

Add $1$ and $1$ .

$56 + 7 \sqrt{2} - 8 \sqrt{2} - {\sqrt{2}}^{2}$

Rewrite ${\sqrt{2}}^{2}$ as $2$ .

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{2}$ as $2^{\frac{1}{2}}$ .

$56 + 7 \sqrt{2} - 8 \sqrt{2} - {(2^{\frac{1}{2}})}^{2}$

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$56 + 7 \sqrt{2} - 8 \sqrt{2} - 2^{\frac{1}{2} \cdot 2}$

Combine $\frac{1}{2}$ and $2$ .

$56 + 7 \sqrt{2} - 8 \sqrt{2} - 2^{\frac{2}{2}}$

Cancel the common factor of $2$ .

Cancel the common factor.

$56 + 7 \sqrt{2} - 8 \sqrt{2} - 2^{\frac{2}{2}}$

Rewrite the expression.

$56 + 7 \sqrt{2} - 8 \sqrt{2} - 2^{1}$

Evaluate the exponent.

$56 + 7 \sqrt{2} - 8 \sqrt{2} - 1 \cdot 2$

Multiply $- 1$ by $2$ .

$56 + 7 \sqrt{2} - 8 \sqrt{2} - 2$

Subtract $2$ from $56$ .

$54 + 7 \sqrt{2} - 8 \sqrt{2}$

Subtract $8 \sqrt{2}$ from $7 \sqrt{2}$ .

$54 - \sqrt{2}$

Step 3

The result can be shown in multiple forms.

Exact Form:

$54 - \sqrt{2}$

Decimal Form:

$52.58578643 \dots$

Simplify (7- square root of 2)(8+ square root of 2)

Related Questions

Support LCMGCF.com by disabling your adblocker.