Question

find value of a and b if 3+4root2/3-4root2=a+b root 2​

Answer 1

$\begin{gathered}\begin{gathered}\bf \: Given - \begin{cases} &\sf{\dfrac{3 + 4 \sqrt{2} }{3 - 4 \sqrt{2} } = a + b \sqrt{2} } \end{cases}\end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered}\bf \: To\:find - \begin{cases} &\sf{the \: value \: of \: a \: and \: b}  \end{cases}\end{gathered}\end{gathered}$

Concept Used :-

Rationalization of the denominator

• So, in order to rationalize the denominator, we need to get rid of all radicals that are in the denominator.

• We just multiply and divide by the conjugate (with opposite sign in between two terms) of the term given in the denominator.

Now,

Let's solve the problem!!

$\large\underline{\bold{Solution :- }}$

$\rm :\longmapsto\:\dfrac{3 + 4 \sqrt{2} }{3 - 4 \sqrt{2} } = a + b \sqrt{2}$

On rationalizing the denominator, we get

$\rm :\implies\:\dfrac{3 + 4 \sqrt{2} }{3 - 4 \sqrt{2} } \times \dfrac{3 + 4 \sqrt{2} }{3 + 4 \sqrt{2} } = a + b \sqrt{2}$

$\rm :\implies\:\dfrac{(3 + 4 \sqrt{2})^{2} }{ {3}^{2} - (4 \sqrt{2})^{2} } = a + b \sqrt{2}$

We know,

$\boxed{ \bf \: {(x + y)}^{2} = {x}^{2} + 2xy + {y}^{2} }$

$\boxed{ \bf \: {x}^{2} - {y}^{2} = (x + y)(x - y)}$

So,

$\rm :\longmapsto\:\dfrac{9 + 32 + 24 \sqrt{2} }{9 - 32} = a + b \sqrt{2}$

$\rm :\longmapsto\:\dfrac{41 + 24 \sqrt{2} }{ - 23} = a + b \sqrt{2}$

$\rm :\longmapsto\: - \dfrac{41}{23} - \dfrac{24}{23} \sqrt{2} = a + b \sqrt{2}$

So, on comparing, we get

$\rm :\implies\:\boxed{ \bf \: a \: = \: - \: \dfrac{41}{23} \: \: \: \: and \: \: \: \: b \: = \: - \: \dfrac{24}{23} }$