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

Asked by Abhisek | 1 year ago |  97

##### Solution :-

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

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 | 1 year ago

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