Solve -2/4+3sqrt{2}

$$\frac{ -2 }{ 4+3 \sqrt{ 2 } }$$

Evaluate

$$4-3\sqrt{2}\approx -0.242640687$$

Solution Steps

Rationalize the denominator of \frac{-2}{4+3\sqrt{2}} by multiplying numerator and denominator by 4-3\sqrt{2}.

$$\frac{-2\left(4-3\sqrt{2}\right)}{\left(4+3\sqrt{2}\right)\left(4-3\sqrt{2}\right)}$$

Consider \left(4+3\sqrt{2}\right)\left(4-3\sqrt{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}.

$$\frac{-2\left(4-3\sqrt{2}\right)}{4^{2}-\left(3\sqrt{2}\right)^{2}}$$

Calculate 4 to the power of 2 and get 16.

$$\frac{-2\left(4-3\sqrt{2}\right)}{16-\left(3\sqrt{2}\right)^{2}}$$

Expand \left(3\sqrt{2}\right)^{2}.

$$\frac{-2\left(4-3\sqrt{2}\right)}{16-3^{2}\left(\sqrt{2}\right)^{2}}$$

Calculate 3 to the power of 2 and get 9.

$$\frac{-2\left(4-3\sqrt{2}\right)}{16-9\left(\sqrt{2}\right)^{2}}$$

The square of \sqrt{2} is 2.

$$\frac{-2\left(4-3\sqrt{2}\right)}{16-9\times 2}$$

Multiply 9 and 2 to get 18.

$$\frac{-2\left(4-3\sqrt{2}\right)}{16-18}$$

Subtract 18 from 16 to get -2.

$$\frac{-2\left(4-3\sqrt{2}\right)}{-2}$$

Cancel out -2 and -2.

$$4-3\sqrt{2}$$