Solve the equation \( x^2 + \dfrac{x}{\sqrt{2}} + 1 = 0\)

Asked by Abhisek | 2 years ago |  144

1 Answer

Solution :-

Given quadratic equation,

\( x^2 + \dfrac{x}{\sqrt{2}} + 1 = 0\)

It can be rewritten as,

\( \sqrt{2}x^2 + x + \sqrt{2} = 0\)

On comparing it with ax2 + bx + c = 0, we have

a = \( \sqrt{2} \), b = 1, and c = \( \sqrt{2} \)

So, the discriminant of the given equation is

D = b2 – 4ac = (1)2 – 4 × \( \sqrt{2} \) × \( \sqrt{2} \) 

= 1 – 8 = -7

Hence, the required solutions are:

\( \dfrac{-b\pm\sqrt{D}}{2a}= \dfrac{-(1)\pm\sqrt{-7}}{2\times \sqrt{2}}\)

\( \dfrac{-1\pm\sqrt{7}i}{2\sqrt{2}}\)

Answered by Pragya Singh | 2 years ago

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