Question

x2-(root3+1)x+root 3=0 solved

Answer 1

Hey there!

Solving it by applying quadratic formula for equation "x^2 - (_/3 + 1)x + _/3 = 0"

For a required quadratic equation of $\bf{ax^2 + bx + c}$ the for which can be represented by the quadratic formula of :

$\boxed{\bf{x_{1, \: 2} = \dfrac{- b +- \sqrt{b^2 - 4ac}}{2a}}}$

Here, a = 1,  b = - _/3,  c = _/3.

Solving for positive and negative values respectively:

$\bf{\dfrac{- (- \sqrt{3} - 1) + \sqrt{(- \sqrt{3} - 1)^2 - 4 \times 1 \times \sqrt{3}}}{2 \times 1}}$

$\bf{\dfrac{- (- 1 - \sqrt{3}) + \sqrt{(1 + \sqrt{3})^2 - 4 \sqrt{3}}}{2}}$

$\bf{\dfrac{- (- 1 - \sqrt{3}) + \sqrt{4 + 2 \sqrt{3} - 4 \sqrt{3}}}{2}}$

$\bf{\dfrac{- (- 1 - \sqrt{3}) + \sqrt{4 - 2 \sqrt{3}}}{2}}$

$\bf{\dfrac{- (- 1 - \sqrt{3}) + \sqrt{4 - 2 \sqrt{3} + \sqrt{3}^2 - 3}{2}}$

$\bf{\dfrac{- (- 1 - \sqrt{3}) + \sqrt{\sqrt{3}^2 - 2 \sqrt{3} + 1}{2}}$

$\bf{\dfrac{- (- 1 - \sqrt{3}) + \sqrt{(\sqrt{3}^2 - \sqrt{3}) + (- \sqrt{3} + 1)}{2}}$

$\bf{\dfrac{- (- 1 - \sqrt{3}) + \sqrt{\sqrt{3} (\sqrt{3} - 1) - (\sqrt{3} - 1)}}{2}}$

$\bf{\dfrac{- (- 1 - \sqrt{3}) + \sqrt{(\sqrt{3} - 1)(\sqrt{3} - 1)}{2}}$

$\bf{\dfrac{- (- 1 - \sqrt{3}) + \sqrt{(\sqrt{3} - 1})^2}{2}}$

Apply the radical rule that is, $\textbf{\sqrt[n]{a^n} = a}$, assuming the value of a is greater than or equal to the value of "0":

$\bf{\dfrac{- (- 1 - \sqrt{3}) + \sqrt{\sqrt{3} - 1}}{2}}$

Expand all the values:

$\bf{\dfrac{\sqrt{3} + 1 + \sqrt{3} - 1}{2}}$

$\bf{\dfrac{2 \sqrt{3}}{2}}$

$\bf{\underline{\therefore \quad x_1 = \sqrt{3}}}$

Similarly find the second value of "x" :

$\bf{\dfrac{- (- \sqrt{3} - 1) + \sqrt{(- \sqrt{3} - 1)^2 - 4 \times 1 \times \sqrt{3}}}{2 \times 1}}$

$\bf{\dfrac{- (- 1 - \sqrt{3}) - \sqrt{(1 + \sqrt{3})^2 - 4 \sqrt{3}}}{2}}$

$\bf{\dfrac{- (- 1 - \sqrt{3}) - \sqrt{4 + 2 \sqrt{3} - 4 \sqrt{3}}}{2}}$

$\bf{\dfrac{- (- 1 - \sqrt{3}) - \sqrt{4 - 2 \sqrt{3}}}{2}}$

$\bf{\dfrac{- (- 1 - \sqrt{3}) - \sqrt{4 - 2 \sqrt{3} + \sqrt{3}^2 - 3}{2}}$

$\bf{\dfrac{- (- 1 - \sqrt{3}) - \sqrt{\sqrt{3}^2 - 2 \sqrt{3} + 1}{2}}$

$\bf{\dfrac{- (- 1 - \sqrt{3}) - \sqrt{(\sqrt{3}^2 - \sqrt{3}) + (- \sqrt{3} + 1)}{2}}$

$\bf{\dfrac{- (- 1 - \sqrt{3}) - \sqrt{\sqrt{3}(\sqrt{3} - 1) - (\sqrt{3} - 1)}{2}}$

$\bf{\dfrac{- (- 1 - \sqrt{3}) - \sqrt{(\sqrt{3} - 1)(\sqrt{3} - 1)}}{2}}$

$\bf{\dfrac{- (- 1 - \sqrt{3}) - \sqrt{(\sqrt{3} - 1}^2}{2}}$

Apply the radical rule that is, $\bf{\sqrt[n]{a^n} = a}$, assuming the value of a is greater than or equal to the value of "0":

$\bf{\dfrac{- (- 1 - \sqrt{3}) - \sqrt{\sqrt{3} - 1}}{2}}$

Expand all the values:

$\bf{\dfrac{\sqrt{3} + 1 - \sqrt{3} + 1}{2}}$

Group the like terms and add similar elements and numbers:

$\bf{\dfrac{2 + \sqrt{3} - \sqrt{3}}{2}}$

$\bf{\dfrac{2}{2}}$

$\bf{\underline{\therefore \quad x_2 = 1}}$

Therefore the final solutions for this quadratic equations are:

$\boxed{\underline{\bf{x_1 = \sqrt{3}}}}$

$\boxed{\underline{\bf{x_2 = 1}}}$

Hope this extremely detailed process helps you and clears the doubts for solving it via quadratic equations!!!!!