Solve frac{sqrt{2}+32}{-6sqrt{2}+2}

$$\frac { \sqrt { 2 } + 32 } { - 6 \sqrt { 2 } + 2 }$$

Evaluate

$$-\frac{97\sqrt{2}}{34}-\frac{19}{17}\approx -5.152315163$$

Solution Steps

Rationalize the denominator of \frac{\sqrt{2}+32}{-6\sqrt{2}+2} by multiplying numerator and denominator by -6\sqrt{2}-2.

$$\frac{\left(\sqrt{2}+32\right)\left(-6\sqrt{2}-2\right)}{\left(-6\sqrt{2}+2\right)\left(-6\sqrt{2}-2\right)}$$

Consider \left(-6\sqrt{2}+2\right)\left(-6\sqrt{2}-2\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$\frac{\left(\sqrt{2}+32\right)\left(-6\sqrt{2}-2\right)}{\left(-6\sqrt{2}\right)^{2}-2^{2}}$$

Expand \left(-6\sqrt{2}\right)^{2}.

$$\frac{\left(\sqrt{2}+32\right)\left(-6\sqrt{2}-2\right)}{\left(-6\right)^{2}\left(\sqrt{2}\right)^{2}-2^{2}}$$

Calculate -6 to the power of 2 and get 36.

$$\frac{\left(\sqrt{2}+32\right)\left(-6\sqrt{2}-2\right)}{36\left(\sqrt{2}\right)^{2}-2^{2}}$$

The square of \sqrt{2} is 2.

$$\frac{\left(\sqrt{2}+32\right)\left(-6\sqrt{2}-2\right)}{36\times 2-2^{2}}$$

Multiply 36 and 2 to get 72.

$$\frac{\left(\sqrt{2}+32\right)\left(-6\sqrt{2}-2\right)}{72-2^{2}}$$

Calculate 2 to the power of 2 and get 4.

$$\frac{\left(\sqrt{2}+32\right)\left(-6\sqrt{2}-2\right)}{72-4}$$

Subtract 4 from 72 to get 68.

$$\frac{\left(\sqrt{2}+32\right)\left(-6\sqrt{2}-2\right)}{68}$$

Apply the distributive property by multiplying each term of \sqrt{2}+32 by each term of -6\sqrt{2}-2.

$$\frac{-6\left(\sqrt{2}\right)^{2}-2\sqrt{2}-192\sqrt{2}-64}{68}$$

The square of \sqrt{2} is 2.

$$\frac{-6\times 2-2\sqrt{2}-192\sqrt{2}-64}{68}$$

Multiply -6 and 2 to get -12.

$$\frac{-12-2\sqrt{2}-192\sqrt{2}-64}{68}$$

Combine -2\sqrt{2} and -192\sqrt{2} to get -194\sqrt{2}.

$$\frac{-12-194\sqrt{2}-64}{68}$$

Subtract 64 from -12 to get -76.

$$\frac{-76-194\sqrt{2}}{68}$$