Question

6-4√2/6+4√2 simplify each of the following by rationalising the denominator

Answer 1

(6-4√2)/(6+4√2) × (6-4√2)/(6-4√2)

= (6-4√2)²/(6²-(4√2)²)

= {6²+(4√2)² - 2(6)(4√2)}/(36-32)

= {36+32-48√2}/4

= (68-48√2)/4

= 68/4 - 48√2/4

= 17 - 12√2

Answer 2

Given: An expression-  $\frac{6 \ - \ 4\sqrt{2} }{6 \ + \ 4\sqrt{2} }$

To find: Simplified form of the given expression by rationalising the denominator

Solution: To rationalise the denominator, first we need to find the rationalisation factor of $6 + 4\sqrt{2}$.

Rationalisation factor is an irrational number that is multiplied with the given irrational expression to make it free from square roots.

Now, the rationalisation factor of $a \ + \ b\sqrt{c}$ will be $a \ - \ b\sqrt{c}$

∵ $(a \ + \ b\sqrt{c} )(a \ - \ b\sqrt{c}) = (a) ^{2} \ - \ (b\sqrt{c}) ^{2} = a^{2} - b^{2} c$, which is rational (free from roots)

Similarly,

∴ $(6 + 4\sqrt{2}) (6 - 4\sqrt{2}) = (6 )^{2} - (4\sqrt{2}) ^{2}$

⇒ $36 - 16(2) = 36 - 32 = 4$, which is rational

∴ Rationalisation factor of  $6 + 4\sqrt{2}$  is  $6 - 4\sqrt{2}$.

Now, $\frac{6 \ - \ 4\sqrt{2} }{6 \ + \ 4\sqrt{2} } = \frac{6 \ - \ 4\sqrt{2} }{6 \ + \ 4\sqrt{2} } \ * \ \frac{6 - 4\sqrt{2}}{6 - 4\sqrt{2}}$

⇒ $\frac{(6 \ - \ 4\sqrt{2})(6 \ - \ 4\sqrt{2}) }{(6 \ + \ 4\sqrt{2})(6 \ - \ 4\sqrt{2})}$

⇒ $\frac{(6-4\sqrt{2}) ^{2} }{4}$     [calculated above]

⇒ $\frac{(6)^{2} \ + \ (4\sqrt{2}) ^{2} \ - \ 2*6*4\sqrt{2} }{4}$

⇒ $\frac{36 \ + \ 16(2) \ - \ 48\sqrt{2} }{4} = \frac{68 \ - \ 48\sqrt{2} }{4}$

⇒ $17 - 12\sqrt{2}$

Hence, simplified form of given expression is $17 - 12\sqrt{2}$.