Question

if one root of the quadratic equation is 1/√4-√3 then quadratic equation can be​

Answer 1

$\underline{\textbf{Given:}}$

$\textsf{One root of a quadratic equation is}\;\mathsf{\dfrac{1}{\sqrt4-\sqrt3}}$

$\underline{\textbf{To find:}}$

$\textsf{The quadratic equation whose one of the root is}\;\mathsf{\dfrac{1}{\sqrt4-\sqrt3}}$

$\underline{\textbf{Solution:}}$

$\mathsf{Take\;x=\dfrac{1}{\sqrt4-\sqrt3}}$

$\mathsf{x=\dfrac{1}{\sqrt4-\sqrt3}{\times}\dfrac{\sqrt4+\sqrt3}{\sqrt4+\sqrt3}}$

$\mathsf{x=\dfrac{\sqrt4+\sqrt3}{(\sqrt4)^2-(\sqrt3)^2}}$

$\mathsf{x=\dfrac{\sqrt4+\sqrt3}{4-3}}$

$\mathsf{x=\dfrac{\sqrt4+\sqrt3}{1}}$

$\mathsf{x=2+\sqrt3}$

$\textsf{This can be written as,}$

$\mathsf{x-2=\sqrt3}$

$\textsf{Squaring on bothsides, we get}$

$\mathsf{(x-2)^2=3}$

$\mathsf{x^2+4-4x=3}$

$\implies\boxed{\bf\,x^2-4x+1=0}$

$\underline{\textbf{Answer:}}$

$\mathsf{The\;required\;quadratic\;equation\;is\;x^2-4x+1=0}$