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

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Question

(A) The given quadratic equation is $x^{2}+x+\frac{1}{\sqrt{2}}=0$. Multiplying by $\sqrt{2}$,we get $\sqrt{2}x^{2}+\sqrt{2}x+1=0$. Comparing with $ax

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) The given quadratic equation is $x^{2}+x+\frac{1}{\sqrt{2}}=0$. Multiplying by $\sqrt{2}$,we get $\sqrt{2}x^{2}+\sqrt{2}x+1=0$. Comparing with $ax^{2}+bx+c=0$,we have $a=\sqrt{2}$,$b=\sqrt{2}$,and $c=1$. The discriminant $D = b^{2}-4ac = (\sqrt{2})^{2}-4(\sqrt{2})(1) = 2-4\sqrt{2}$. Since $D $x = \frac{-\sqrt{2} \pm i\sqrt{4\sqrt{2}-2}}{2\sqrt{2}} = \frac{-\sqrt{2} \pm i\sqrt{2(2\sqrt{2}-1)}}{2\sqrt{2}}$. $x = \frac{-\sqrt{2} \pm i\sqrt{2}\sqrt{2\sqrt{2}-1}}{2\sqrt{2}} = \frac{-1 \pm i\sqrt{2\sqrt{2}-1}}{2}$.

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

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