Find the value of the following (explain too)
[tex]\frac{\sqrt{3} +1}{\sqrt{3} -1} = a+b\sqrt{3}[/tex][tex]\frac{\sqrt{3} +1

1 thought on “Find the value of the following (explain too)<br /> [tex]\frac{\sqrt{3} +1}{\sqrt{3} -1} = a+b\sqrt{3}[/tex][tex]\frac{\sqrt{3} +1”

1. [tex]\huge \fbox \blue{Answer★}[/tex]

   We Have,

   [tex] = \frac{ \sqrt{3} + 1 }{ \sqrt{3} – 1 } – \frac{ \sqrt{3} – 1}{ \sqrt{3} – 1 } = a + b \sqrt{3} [/tex]

   [tex]⇒ \frac{( \sqrt{3} + 1 {)}^{2} – ( \sqrt{3} – 1 {)}^{2} }{( \sqrt{3} – 1)( \sqrt{3} + 1) } = a + b \sqrt{3} [/tex]

   [tex]\[ \left[ \text{Using Formula = } {a}^{2} – {b}^{2} = (a + b)(a – b)\right] \] [/tex]

   [tex]⇒ \frac{( \sqrt{3} + 1 + \sqrt{3} – 1)( \sqrt{3} + 1 – \sqrt{3} + 1) }{(( \sqrt{3} {)}^{2} – {1}^{2} ) } [/tex]

   [tex]⇒ \frac{2 \sqrt{3} \times 2 }{3 – 1} = a + b \sqrt{3} [/tex]

   [tex]⇒2 \sqrt{3} = a + b \sqrt{3} [/tex]

   [tex] \textbf{Comparing L.H.S Ans R.H.S we get,} [/tex]

   [tex]b = 2 \: \textbf{and} \: a = 0[/tex]

   Thus Value of a and b are 0 and 2 respectively.

   [tex] \\ \\ \\ \\ \\ \\ \\ \\ \\ \sf \colorbox{gold} {\red★ANSWER ᵇʸ ⁿᵃʷᵃᵇ⁰⁰⁰⁸}[/tex]

   Reply