Question

The set of values of k for which the given quadratic equation has real roots $2{x^2} + kx + 3 = 0$ is $k \leqslant - 2\sqrt 6 $.
$
  \left( A \right){\text{True}} \\
  \left( B \right){\text{False}} \\
$

Answer 1

Hint-In this question, we use the concept of nature of roots. For real roots discriminants are always greater than equal to zero. Let’s take a quadratic equation \[a{x^2} + bx + c = 0\] then the value of discriminant, $D = {b^2} - 4ac$ and for real roots $D \geqslant 0$.

Complete step-by-step solution -
Given, a quadratic equation $2{x^2} + kx + 3 = 0$
Now, we compare the given quadratic equation $2{x^2} + kx + 3 = 0$ with the general quadratic equation \[a{x^2} + bx + c = 0\] to find the coefficients a, b and c .
Then after comparing, a=2, b=k and c=3.
Now, using the concept of the nature of roots .
For real roots, $D \geqslant 0$ and value of $D = {b^2} - 4ac$.
$ \Rightarrow {b^2} - 4ac \geqslant 0$
Put the value of a, b and c.
$
   \Rightarrow {k^2} - 4 \times 2 \times 3 \geqslant 0 \\
   \Rightarrow {k^2} - 24 \geqslant 0 \\
   \Rightarrow {k^2} - {\left( {2\sqrt 6 } \right)^2} \geqslant 0 \\
   \Rightarrow \left( {k + 2\sqrt 6 } \right)\left( {k - 2\sqrt 6 } \right) \geqslant 0 \\
$
Suppose $k - 2\sqrt 6 $ and $k + 2\sqrt 6 $ are both positive. But $k + 2\sqrt 6 $ is bigger than $k - 2\sqrt 6 $, so it is enough for $k - 2\sqrt 6 $ to be positive; if it is, then the other one will be too. $k - 2\sqrt 6 $ is positive whenever $k \geqslant 2\sqrt 6 $.
Or both $k - 2\sqrt 6 $ and $k + 2\sqrt 6 $ could be negative. But $k + 2\sqrt 6 $ is bigger than $k - 2\sqrt 6 $, so if $k + 2\sqrt 6 $ is negative, $k - 2\sqrt 6 $ is also. And $k + 2\sqrt 6 $ is negative whenever $k \leqslant -2\sqrt 6 $
So, the set of values of k is $ - 2\sqrt 6 \leqslant k \leqslant 2\sqrt 6 $ .
Hence, the correct option is (b).

Note-In such types of questions we generally face the problem of solving the inequality equation. So, the better way to solve inequality we use a wavy curve method in which we represent the critical point on the number line and then put any number on the equation and check if the sign of the equation is positive or negative and then make an interval according to the equation.