Solve frac{sqrt{4+3}}{sqrt{6}+1}=

$$\frac { \sqrt { 4 + 3 } } { \sqrt { 6 } + 1 } =$$

Evaluate

$$\frac{\sqrt{7}\left(\sqrt{6}-1\right)}{5}\approx 0.766997877$$

Solution Steps

Add 4 and 3 to get 7.

$$\frac{\sqrt{7}}{\sqrt{6}+1}$$

Rationalize the denominator of \frac{\sqrt{7}}{\sqrt{6}+1} by multiplying numerator and denominator by \sqrt{6}-1.

$$\frac{\sqrt{7}\left(\sqrt{6}-1\right)}{\left(\sqrt{6}+1\right)\left(\sqrt{6}-1\right)}$$

Consider \left(\sqrt{6}+1\right)\left(\sqrt{6}-1\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$\frac{\sqrt{7}\left(\sqrt{6}-1\right)}{\left(\sqrt{6}\right)^{2}-1^{2}}$$

Square \sqrt{6}. Square 1.

$$\frac{\sqrt{7}\left(\sqrt{6}-1\right)}{6-1}$$

Subtract 1 from 6 to get 5.

$$\frac{\sqrt{7}\left(\sqrt{6}-1\right)}{5}$$

Use the distributive property to multiply \sqrt{7} by \sqrt{6}-1.

$$\frac{\sqrt{7}\sqrt{6}-\sqrt{7}}{5}$$

To multiply \sqrt{7} and \sqrt{6}, multiply the numbers under the square root.

$$\frac{\sqrt{42}-\sqrt{7}}{5}$$