Question

quadratic equation x^2 - 2x +5 =0

Answer 1

Hey there!

Solving it by applying quadratic formula for equation "x^2 - 2x + 5 = 0"

For a required quadratic equation of $\bf{ax^2 + bx + c = 0}$ the solutions 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 = - 2,  c = 5.

Solving for positive and negative values respectively:

$\bf{x_{1} = \dfrac{- (- 2) + \sqrt{(- 2)^2 - 4(1)(5)}}{2(1)}}$

By Applying imaginary number rule that is $\bf{\sqrt{- 1} = i}$.

$\bf{x_{1} = \dfrac{2 + \sqrt{16} i}{2 \times 1}}$

$\bf{x_{1} = \dfrac{2 + 4 i}{2 \times 1}}$

$\bf{x_{1} = \dfrac{2 + 4 i}{2}}$

$\bf{x_{1} = \dfrac{2(1 + 2 i)}{2}}$

$\bf{x_{1} = (1 + 2 i)}$

$\bf{\therefore \quad x_{1} = 1 + 2 i}$

For the second solution in negative form of equation:

$\bf{x_{2} = \dfrac{- (- 2) - \sqrt{(- 2)^2 - 4(1)(5)}}{2(1)}}$

By Applying imaginary number rule that is $\bf{\sqrt{- 1} = i}$.

$\bf{x_{2} = \dfrac{2 - \sqrt{16} i}{2 \times 1}}$

$\bf{x_{2} = \dfrac{2 - 4 i}{2 \times 1}}$

$\bf{x_{2} = \dfrac{2 - 4 i}{2}}$

$\bf{x_{2} = \dfrac{2(1 - 2 i)}{2}}$

$\bf{x_{2} = (1 - 2 i)}$

$\bf{\therefore \quad x_{2} = 1 - 2 i}$

Therefore the final solutions for this quadratic equations are:

$\boxed{\underline{\bf{x_{1} = 1 + 2 i}}}$

$\boxed{\underline{\bf{x_{2} = 1 - 2 i}}}$

Hope this helps you and solves the doubts for getting the equation solved by quadratic method!