Solve 4-45/49-8sqrt{5}

$$\frac { 4 - 45 } { 49 - 8 \sqrt { 5 } }$$

Evaluate

$$\frac{-328\sqrt{5}-2009}{2081}\approx -1.317842526$$

Solution Steps

Subtract 45 from 4 to get -41.

$$\frac{-41}{49-8\sqrt{5}}$$

Rationalize the denominator of \frac{-41}{49-8\sqrt{5}} by multiplying numerator and denominator by 49+8\sqrt{5}.

$$\frac{-41\left(49+8\sqrt{5}\right)}{\left(49-8\sqrt{5}\right)\left(49+8\sqrt{5}\right)}$$

Consider \left(49-8\sqrt{5}\right)\left(49+8\sqrt{5}\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$\frac{-41\left(49+8\sqrt{5}\right)}{49^{2}-\left(-8\sqrt{5}\right)^{2}}$$

Calculate 49 to the power of 2 and get 2401.

$$\frac{-41\left(49+8\sqrt{5}\right)}{2401-\left(-8\sqrt{5}\right)^{2}}$$

Expand \left(-8\sqrt{5}\right)^{2}.

$$\frac{-41\left(49+8\sqrt{5}\right)}{2401-\left(-8\right)^{2}\left(\sqrt{5}\right)^{2}}$$

Calculate -8 to the power of 2 and get 64.

$$\frac{-41\left(49+8\sqrt{5}\right)}{2401-64\left(\sqrt{5}\right)^{2}}$$

The square of \sqrt{5} is 5.

$$\frac{-41\left(49+8\sqrt{5}\right)}{2401-64\times 5}$$

Multiply 64 and 5 to get 320.

$$\frac{-41\left(49+8\sqrt{5}\right)}{2401-320}$$

Subtract 320 from 2401 to get 2081.

$$\frac{-41\left(49+8\sqrt{5}\right)}{2081}$$

Use the distributive property to multiply -41 by 49+8\sqrt{5}.

$$\frac{-2009-328\sqrt{5}}{2081}$$