Question

if 3+2root 3 / 3-root 3 = a+root 3b then the value of a+b where a nd b are rational numbers is?

Answer 1

Answer:

a= 5/2 , b= 3/2

Step-by-step explanation:

=> 3+2√3  = a+√3 b

      3-√3

=> 3+2√3 × 3+√3 = 9+3√3+6√3+6 = 15+9√3 = 5+3√3

      3-√3         3+√3             6                      6         2   2

=>  comparing with a+b√3,

      we have , a= 5/2    b=3/2

   

Answer 2

Answer:

$\bigstar{\bold{Value\:of\:a=\dfrac{5}{2} }}$

$\bigstar{\bold{Value\:of\:b=\dfrac{3}{2} }}$

Step-by-step explanation:

$\Large{\underline{\underline{\bf{Given:}}}}$

$\sf{\dfrac{3+2\sqrt{3} }{3-\sqrt{3} }=a+b\sqrt{3} }$

$\Large{\underline{\underline{\bf{To\:Find:}}}}$

• The values of a and b

$\Large{\underline{\underline{\bf{Solution:}}}}$

→ Given,

  $\sf{\dfrac{3+2\sqrt{3} }{3-\sqrt{3} } }$

→ Rationalising the denominator by multiplying 3 + √3 on both numerator and denominator

   $\sf{\dfrac{3+2\sqrt{3}\times (3+\sqrt{3}) }{3-\sqrt{3}\times (3+\sqrt{3} ) } }$

→ Simplifying by using identities,

  $\sf{\dfrac{9+3\sqrt{3}+6\sqrt{3} +6 }{3^{2} -(\sqrt{3})^{2} } }$

$\sf{=\dfrac{9+9\sqrt{3}+6 }{9-3} }$

→ Taking 3 common from the numerator,

  $\sf{=\dfrac{3(3+3\sqrt{3}+2) }{6} }$

→ Cancelling the numerator and denominator by 3

   $\sf{=\dfrac{(3+3\sqrt{3}+2) }{2} }$

  $\sf{=\dfrac{3}{2}+\dfrac{3\sqrt{3} }{2}+\dfrac{2}{2} }$

 = 5/2 + 3/2 ×√3

→ Equating it we get the value of a as 5/2 and b as 3/2

$\boxed{\bold{Value\:of\:a=\dfrac{5}{2} }}$

$\boxed{\bold{Value\:of\:b=\dfrac{3}{2} }}$

$\Large{\underline{\underline{\bf{Notes:}}}}$

(a + b)² = a² + 2ab + b²

(a - b)² = a² - 2ab + b²

a² - b²= (a + b)  (a - b)