Question

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

Answer 1

GIVEN:

• 6 - 4√2/6 + 4√2 = a + b√2

TO FIND:

• a , b

SOLUTION:

6 - 4√2/6 + 4√2 = a + b√2

Taking LHS and simplifying by rationalizing th denominator

→ (6 - 4√2)(6 - 4√2)/(6 + 4√2)(6 - 4√2)

Using

♦ (a - b)(a - b) = (a - b)² = a² - 2ab + b²

♦ (a + b)(a - b) = a² - b²

→ 6² - 2(4√2)(6) + (4√2)²/6² - (4√2)²

→ 36 - 48√2 + 32/36 - 32

→ 68 - 48√2/4

Taking common

→ 4(17 - 12√2)/4

→ 17 - 12√2

→ 17 + (-12)(√2)

Now, comparing with RHS

→ 17 + (- 12√2) = a + b√2

a = 17 [.Ans]

b = - 12. [.Ans]

Answer 2

Given

$\bf Simplify \:\: \dfrac{6-4\sqrt{2}}{6+4\sqrt{2}}=a+b\sqrt{2}$

Solution:

$\bf \dfrac{6-4\sqrt{2}}{6+4\sqrt{2}}\times \dfrac{6-4\sqrt{2}}{6-4\sqrt{2}}\:\:$

$\bf = \dfrac{(6-4\sqrt{2})^{2}}{(6)^{2} - (4\sqrt{2})^{2}}$

[ ∴ (a+b)(a-b) = (a)² - (b)² ]

$\bf = \dfrac{(6)^{2} + (4\sqrt{2})^{2} - 2 \times (6) \times (4\sqrt{2} )}{36 - 32}$

[ ∴ (a - b)² = (a)² + (b)² - 2ab ]

$\bf = \dfrac{36 + 32 - 48\sqrt{2}}{4}$

$\bf = \dfrac{68 - 48\sqrt{2}}{4}$

$\bf = \dfrac{ \cancel{4} (17 -12\sqrt{2})}{ \cancel{4} }$

[Taking 4 as common ]

$\bf = 17 - 12\sqrt{2} = a+b\sqrt{2}$

so,

$\bf = 17 + (-12\sqrt{2}) = a+b\sqrt{2}$

a = 17

And

b = -12