Solve the equation x^{2}+frac{x}{sqrt{2}}+1=0 | vedclass

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Question

(A) The given quadratic equation is $x^{2}+\frac{x}{\sqrt{2}}+1=0$. Multiply the entire equation by $\sqrt{2}$ to simplify: $\sqrt{2}x^{2}+x+\sqrt{2}=

Vedclass · Accepted Answer

$\frac{-1 \pm \sqrt{7} i}{2 \sqrt{2}}$

Vedclass · Answer

(A) The given quadratic equation is $x^{2}+\frac{x}{\sqrt{2}}+1=0$. Multiply the entire equation by $\sqrt{2}$ to simplify: $\sqrt{2}x^{2}+x+\sqrt{2}=0$. Comparing this with the standard quadratic form $ax^{2}+bx+c=0$,we get $a=\sqrt{2}$,$b=1$,and $c=\sqrt{2}$. The discriminant $D$ is given by $D = b^{2}-4ac = (1)^{2}-4(\sqrt{2})(\sqrt{2}) = 1-8 = -7$. The roots are given by the quadratic formula $x = \frac{-b \pm \sqrt{D}}{2a}$. Substituting the values,we get $x = \frac{-1 \pm \sqrt{-7}}{2\sqrt{2}} = \frac{-1 \pm \sqrt{7}i}{2\sqrt{2}}$ (since $\sqrt{-1}=i$).

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

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