Question

if 3+ √7/ 3- √7 = a + b√7, find a and b.

Answer 1

GIVEN :-

• 3 +√7 /3 -√7 = a + b√7

TO FIND :-

• The value of a and b

SOLUTION :-

⇒ (3 + √7)/(3 - √7) = a + b√7

⇒ (3 + √7)/(3 - √7) × (3 + √7)/(3 + √7) = a + b√7

⇒ (3 + √7)²/(3)³ - (√7)² = a + b√7

⇒ [3² + (√7)² + 2 × 3 × √7]/9 - 7 = a + b√7

⇒ (9 + 7 + 6√7)/2 = 4 + b√7

⇒ (16 + 6√7)/2 = a + b√7

⇒ (16/2) + (6√7/2) = a + b√7

⇒ (16/2) + (6/2)√7 = a + b√7

On comparing both side ,

⇒ a = 16/2 = 8

⇒ b = 6/2 = 3

Hence the value of a is 8 and b is 3.

Answer 2

Answer

• a = 8 , b = 3

Given

$\bullet \;\; \rm \dfrac{3+\sqrt{7}}{3-\sqrt{7}}=a+b\sqrt{7}$

To Find

• Value of a and b

Formula Used

$\boxed{\begin{minipage}{4cm} \bullet \;\; \rm (x+y)(x-y)=x^2-y^2\\\\\bullet \;\; \rm (x+y)^2=x^2+y^2+2xy\end{minipage}}$

Solution

$\rm \dfrac{3+\sqrt{7}}{3-\sqrt{7}}=a+b\sqrt{7}$

Rationalizing the denominator , we get ,

$\implies \rm \dfrac{3+\sqrt{7}}{3-\sqrt{7}}\times \dfrac{3+\sqrt{7}}{3+\sqrt{7}}=a+b\sqrt{7}\\\\\implies \rm \dfrac{(3+\sqrt{7})^2}{3^2-(\sqrt{7})^2}=a+b\sqrt{7}\\\\\implies \rm \dfrac{3^2+(\sqrt{7})^2+2.3.\sqrt{7}}{9-7}=a+b\sqrt{7}\\\\\implies \rm \dfrac{9+7+6\sqrt{7}}{2}=a+b\sqrt{7}\\\\\implies \rm \dfrac{16+6\sqrt{7}}{2}=a+b\sqrt{7}\\\\\implies \rm \dfrac{2(8+3\sqrt{7})}{2}=a+b\sqrt{7}\\\\\implies \rm 8+3\sqrt{7}=a+b\sqrt{7}\\\\\implies \rm a=8\ ,b\sqrt{7}=3\sqrt{7}\\\\\implies \rm a=8\ ,b=3$