Question

4+√2/2-√2 = a-√b find value of a and b​

Answer 1

Given

$\sf \: \dfrac{4 + \sqrt{2} }{2 + \sqrt{2} } = a - \sqrt{b}$

To Find

• Value of a and b

Solution

$\sf \: \dfrac{4 + \sqrt{2} }{2 + \sqrt{2} }$

• Multiply numerator and denominator by a radical that will get rid of the radical in the denominator.

$\implies \sf \: \dfrac{4 + \sqrt{2} }{2 + \sqrt{2} } \times \sf \: \dfrac{2 - \sqrt{2} }{2 - \sqrt{2} }$

$\implies \sf \: \dfrac{(4 + \sqrt{2})(2 - \sqrt{2} ) }{2 ^{2} + (\sqrt{2} )^{2} }$

$\implies \sf \: \dfrac{6 - 2 \sqrt{2} }{4 - 2 }$

$\implies \sf \: \dfrac{2(3 - \sqrt{2}) }{ 2 }$

$\implies \sf \: 3 - 2 \sqrt{2} = a - \sqrt{b}$

• $\boxed{\sf a = 3 , b = 2}$

Answer 2

Answer:

Step-by-step explanation:

Given,

4 + √2 / 2 - √2 = a - √b

= Now factorize the denominator with 2 + √2,

= (4 + 2√2 / 2- √2) × (2 + √2 / 2 + √2)

= (4 + 2√2 × 2 + √2) / (2 - √2 × 2 +√2)

= (4 ×2 + 4 × √2 + 2√2 × 2 + 2√2 ×√2) / [(2)^2 - (√2)^2] [∵(a + b) (a - b ) = a ^2 - b^2]

= (8 + 4√2 + 4√2 + 2 (√2)^2 ) / (4 - 2) [∵(√2)^2 = root and square are cancelled ]

= (8 + 8√2 + 2 × 2) / 2 [∵(√2)^2 = root and square are cancelled ]

= (8 + 8√2 + 4 ) / 2

= (12 + 8√2 ) / 2

= 4 (3 + 2√2) / 2

= 2 (3 + 2√2) (∵numerator 4 and denominator 2 are cancelled, we get 2)

= 6 + 4√2 is the answer.