Question:

If $ {{x}^{2}}+4ax+2>0 $ for all values of $ x, $ then $a$ lies in the interval

Updated On: Apr 8, 2024
  • $ (-2,\text{ }4) $
  • $(1, 2)$
  • $ (-\sqrt{2},\sqrt{2}) $
  • $ \left( -\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}} \right) $
Hide Solution
collegedunia
Verified By Collegedunia

The Correct Option is D

Solution and Explanation

$ {{x}^{2}}+4ax+2>0 $
$ \therefore $ $ {{(4a)}^{2}}-4{{x}^{2}}<0 $
[ $ \because $ if $ f(x)>0, $ then $ D<0 $ ]
$ \Rightarrow $ $ 16{{a}^{2}}<8 $
$ \Rightarrow $ $ {{a}^{2}}
Was this answer helpful?
0
0

Concepts Used:

Complex Numbers and Quadratic Equations

Complex Number: Any number that is formed as a+ib is called a complex number. For example: 9+3i,7+8i are complex numbers. Here i = -1. With this we can say that i² = 1. So, for every equation which does not have a real solution we can use i = -1.

Quadratic equation: A polynomial that has two roots or is of the degree 2 is called a quadratic equation. The general form of a quadratic equation is y=ax²+bx+c. Here a≠0, b and c are the real numbers.