Solve sqrt{5} x^{2} + x + sqrt{5} = 0. | vedclass

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

Question

(A) Given the quadratic equation $\sqrt{5} x^{2} + x + \sqrt{5} = 0$. Comparing this with $ax^{2} + bx + c = 0$,we have $a = \sqrt{5}$,$b = 1$,and $c

Vedclass · Accepted Answer

$\frac{-1 \pm \sqrt{19} i}{2 \sqrt{5}}$

Vedclass · Answer

(A) Given the quadratic equation $\sqrt{5} x^{2} + x + \sqrt{5} = 0$. Comparing this with $ax^{2} + bx + c = 0$,we have $a = \sqrt{5}$,$b = 1$,and $c = \sqrt{5}$. The discriminant $D = b^{2} - 4ac$ is calculated as: $D = (1)^{2} - 4(\sqrt{5})(\sqrt{5}) = 1 - 4(5) = 1 - 20 = -19$. The roots are given by the quadratic formula $x = \frac{-b \pm \sqrt{D}}{2a}$. Substituting the values,we get $x = \frac{-1 \pm \sqrt{-19}}{2\sqrt{5}}$. Since $\sqrt{-1} = i$,the solutions are $x = \frac{-1 \pm \sqrt{19} i}{2\sqrt{5}}$.

Solve $\sqrt{5} x^{2} + x + \sqrt{5} = 0$.

Similar Questions

If the roots of the equation $x^2 - 8x + (a^2 - 6a) = 0$ are real,then:

Let one root of $ax^2 + bx + c = 0$,where $a, b, c$ are integers,be $3 + \sqrt{5}$. Then the other root is:

The equation $\sqrt{3x^2 + x + 5} = x - 3$,where $x$ is real,has

If $\alpha$ and $\beta$ are the roots of the equation $2^{6x} - 3(2^{3x+2}) + 32 = 0$ with $\beta < 1$,then $2\alpha + 3\beta =$

What is the sum of all natural numbers $n$ such that the product of the digits of $n$ (in base $10$) is equal to $n^2-10n-36$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module