Question

Find the value of a and b​

Answer 1

Answer:

a=3 and b=2

Step-by-step explanation:

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Answer 2

Question:

Find the value of a and b.

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

Answer:

$\displaystyle{{\boxed{\red{\sf\:a\:=\:-\:7\:}}}\:\:\&\:\:\:{\boxed{\red{\sf\:b\:=\:-\:4\:}}}}$

Step-by-step-explanation:

The given equation is

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

We have to find the value of a and b.

Now,

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

$\displaystyle{\implies\sf\:LHS\:=\:\dfrac{3\:+\:2\:\sqrt{3}}{3\:-\:2\:\sqrt{3}}}$

$\displaystyle{\implies\sf\:LHS\:=\:\dfrac{3\:+\:2\:\sqrt{3}}{3\:-\:2\:\sqrt{3}}\:\times\:\dfrac{3\:+\:2\:\sqrt{3}}{3\:+\:2\:\sqrt{3}}}$

$\displaystyle{\implies\sf\:LHS\:=\:\dfrac{(\:3\:+\:2\:\sqrt{3}\:)^2}{(\:3\:)^2\:-\:(\:2\:\sqrt{3}\:)^2}\:\quad\:-\:-\:-\:[\:\because\:a^2\:-\:b^2\:=\:(\:a\:+\:b\:)\:(\:a\:-\:b\:)\:]}$

$\displaystyle{\implies\sf\:LHS\:=\:\dfrac{3^2\:+\:2\:\times\:3\:\times\:2\:\sqrt{3}\:+\:(\:2\:\sqrt{3}\:)^2}{9\:-\:4\:\times\:3}\:\quad\:-\:-\:-\:[\:\because\:(\:a\:+\:b\:)^2\:=\:a^2\:+\:2ab\:+\:b^2\:]}$

$\displaystyle{\implies\sf\:LHS\:=\:\dfrac{9\:+\:12\:\sqrt{3}\:+\:4\:\times\:3}{9\:-\:12}}$

$\displaystyle{\implies\sf\:LHS\:=\:\dfrac{9\:+\:12\:\sqrt{3}\:+\:12}{-\:3}}$

$\displaystyle{\implies\sf\:LHS\:=\:-\:\dfrac{9\:+\:12\:+\:12\:\sqrt{3}}{3}}$

$\displaystyle{\implies\sf\:LHS\:=\:-\:\dfrac{21\:+\:12\:\sqrt{3}}{3}}$

$\displaystyle{\implies\sf\:LHS\:=\:-\:\left(\:\cancel{\dfrac{21}{3}}\:+\:\dfrac{\cancel{12}\:\sqrt{3}}{\cancel{3}}\:\right)}$

$\displaystyle{\implies\sf\:LHS\:=\:-\:(\:7\:+\:4\:\sqrt{3}\:)}$

$\displaystyle{\implies\sf\:LHS\:=\:-\:7\:-\:4\:\sqrt{3}}$

$\displaystyle{\implies\sf\:RHS\:=\:a\:+\:b\:\sqrt{3}}$

$\displaystyle{\implies\sf\:a\:=\:-\:7\:\:\&\:\:b\:=\:-\:4\:\:\quad\:-\:-\:-\:[\:By\:comparing\:]}$

$\displaystyle{\therefore\:\underline{\boxed{\red{\sf\:a\:=\:-\:7\:}}}\:\:\:\&\:\:\:\underline{\boxed{\red{\sf\:b\:=\:-\:4\:}}}}$