# Extra Questions For CBSE Class 9 Maths Chapter 1 | Q6

###### Number Systems | Operation on Real numbers - Rationalize the denominators

Question 6: Rationalize the denominator
=$$frac{1}{8 + 3\sqrt{5}}$× $\frac{8 - 3\sqrt{5}}{8 - 3\sqrt{5}}$ = $\frac{8 - 3\sqrt{5}}{8^2 -$3$sqrt{5}$^2}) = $\frac{8 - 3\sqrt{5}}{64 - 45}$ = $\frac{8 - 3\sqrt{5}}{19}$ ####$d) $$frac{1}{4 + \sqrt{2} + \sqrt{5}}$ This one is more difficult than the previous 3 questions. Let us do it in 2 steps. Step 1: Multiply and divide the expression by$4 – √2 + √5)
$$frac{1}{4 + \sqrt{2} + \sqrt{5}}$ × $\frac{4 -$$sqrt{2} + \sqrt{5}$}{4 -$$sqrt{2} + $sqrt{5}$}) = $\frac{4 - \sqrt{2} - \sqrt{5}}{4^2 - \sqrt{2} + \sqrt{5}^2}$ = $\frac{4 - \sqrt{2} - \sqrt{5}}{16 -$2 + 5 + 2$sqrt{10}$}) = $\frac{4 - \sqrt{2} - \sqrt{5}}{9 - 2\sqrt{10}}$ Step 2: Multiply and divide the expression by the conjugate of the denominator. i.e., by$9 + 2√10)
$\frac{4 - \sqrt{2} - \sqrt{5}}{9 - 2\sqrt{10}}$ × $\frac{9 + 2\sqrt{10}}{9 + 2\sqrt{10}}$

= $\frac{36 + 8\sqrt{10} - 9\sqrt{2} - 2\sqrt{20} - 9\sqrt{5} - 2\sqrt{50}}{81 - 40}$

= $\frac{36 + 8\sqrt{10} - 9\sqrt{2} - 4\sqrt{5} - 9\sqrt{5} - 10\sqrt{2}}{41}$

= $\frac{36 + 8\sqrt{10} - 19\sqrt{2} - 13\sqrt{5}}{41}$

###### Free CBSE Online CoachingClass 9 Maths

Register in 2 easy steps and
Start learning in 5 minutes!