Question

find the value of a-√b​

Answer 1

Question:

Find the value of a & b

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

Answer:

a = 3 & b = 2

Step-by-step explanation:

Rationalizing factor = 2 - √2

Multiply and divide the given fraction by (2 - √2)

  $\sf =\dfrac{4+\sqrt{2}}{2+\sqrt{2}} \\\\\\ \sf =\dfrac{4+\sqrt{2}}{2+\sqrt{2}} \times \dfrac{2-\sqrt{2}}{2-\sqrt{2}} \\\\\\ \sf = \dfrac{(4+\sqrt{2})(2-\sqrt{2})}{(2+\sqrt{2})(2-\sqrt{2})} \\\\\\ \sf = \dfrac{4(2-\sqrt{2})+\sqrt{2}(2-\sqrt{2})}{2(2-\sqrt{2})+\sqrt{2}(2-\sqrt{2})} \\\\\\ \sf = \dfrac{8-4\sqrt{2}+2\sqrt{2}-\sqrt{2}^2}{4-2\sqrt{2}+2\sqrt{2}-\sqrt{2}^2} \\\\\\ \sf = \dfrac{8-2\sqrt{2}-2}{4-2} \\\\\\ \sf = \dfrac{6-2\sqrt{2}}{2} \\\\\\ \sf =3-\sqrt{2}$

Comparing 3 - √2 with a - √b , we get

a = 3

b = 2

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★ Rationalizing factor :

⇒ The factor of multiplication by which rationalization is done, is called as rationalizing factor.

⇒ If the product of two surds is a rational number, then each surd is a rationalizing factor to other.

⇒ To find the rationalizing factor,  

     =>  If the denominator contains 2 terms, just change the sign between the two terms.

         For example, rationalizing factor of (3 + √2) is (3 - √2)

     => If the denominator contains 1 term, the radical found in the denominator is the factor.

         For example, rationalizing factor of √2 is √2

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Answer 2

$\large \mathfrak{ \color{navy}{⬗ \: Qu}} \mathfrak{ \purple{est}} \mathfrak{ \pink{ion}}$

Find the value of a & b :

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

$\large \mathfrak{ \color{navy}{An}} \mathfrak{ \purple{sw}} \mathfrak{ \pink{er}}$

a = 3 & b = 2

$\large \mathfrak{ \color{royalblue}{Step}} \mathfrak{ \color{navy}{-by-}} \mathfrak{ \purple{step}} \mathfrak{ \pink{ \: Explanation}}$

• Rationalizing factor = 2 - √2

• Multiply and divide the given fraction by (2 - √2)

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

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

$\sf = \dfrac{(4+\sqrt{2})(2-\sqrt{2})}{(2+\sqrt{2})(2-\sqrt{2})}$

$\sf = \dfrac{4(2-\sqrt{2})+\sqrt{2}(2-\sqrt{2})}{2(2-\sqrt{2})+\sqrt{2}(2-\sqrt{2})}$

$\sf = \dfrac{8-4\sqrt{2}+2\sqrt{2}-\sqrt{2}^2}{4-2\sqrt{2}+2\sqrt{2}-\sqrt{2}^2}$

$\sf = \dfrac{8-2\sqrt{2}-2}{4-2}$

$\sf = \dfrac{6-2\sqrt{2}}{2}$

$\sf =3-\sqrt{2}$

• Comparing 3 - √2 with a - √b , we get

a = 3

b = 2

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