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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

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

Similar Questions

To remove the second term of the equation $x^4-8x^3+x^2-x+3=0$,diminish the roots of the equation by

Statement-$I$: If $a + b + c = 0$ and $a, b, c$ are rational,then the roots of the equation $(b + c - a)x^2 + (c + a - b)x + (a + b - c) = 0$ are rational.
Statement-$II$: The discriminant of $(b + c - a)x^2 + (c + a - b)x + (a + b - c) = 0$ is a perfect square.

If $x = 2 + \sqrt{3}$,then what is the value of $x^3 - 7x^2 + 13x - 12$?

The number of real solutions of the equation $|x|^2 - 3|x| + 2 = 0$ is

If $f(x)=2 x^4-13 x^2+a x+b$ is divisible by $x^2-3 x+2$,then $(a, b)$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module