Question

quadratic equation 2 x square minus under root 5 x + 1 = 0

Answer 1

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Answer 2

Answer:

$x\ =\ \dfrac{-\sqrt{5}\ \pm\ 3i}{4}$

Step-by-step explanation:

Given,

The quadratic equation is $2x^2\ -\ \sqrt{5}x\ +\ 1\ =\ 0$

Now the roots of the quadratic equation,

$x\ =\ \dfrac{-b\ \pm\ \sqrt{b^2\ -\ 4ac}}{2a}$

Now

• a = 2
• b = $-\sqrt{5}$
• c = 1

Substituting these values in the above expression of the roots of the quadratic equation,

$x\ =\ \dfrac{-\sqrt{5}\ \pm\ \sqrt{\sqrt{5}^2\ -\ 4\times 2\times 1}}{2\times 2}\\\Rightarrow x\ =\ \dfrac{-\sqrt{5}\ \pm\ \sqrt{-3}}{4}$

Hence the discriminant becomes negative, therefore the roots are in the pair of complex number,

$\therefore x\ =\ \dfrac{-\sqrt{5}\ \pm\ 3i}{4}$