Question

How do you find the discriminant of $7{x^2} + 6x + 2 = 0$ and use it to determine if the equation has one, two real or two imaginary roots?

Answer 1

Hint: This problem deals with finding the determinant and solving the given quadratic equation. If the quadratic equation is in the form of $a{x^2} + bx + c = 0$, then we know that the roots of this quadratic equation are given by :
$ \Rightarrow x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
Here the determinant is given by: ${b^2} - 4ac$. If the discriminant is less than zero, then the roots are imaginary.
If ${b^2} - 4ac < 0$, then the quadratic equation has no real roots but has complex conjugate roots.

Complete step-by-step solution:
Here consider the given quadratic equation $7{x^2} + 6x + 2 = 0$, as given below;
Now comparing the given quadratic equation with standard form $a{x^2} + bx + c = 0$, the coefficients are:
$ \Rightarrow a = 7,b = 6$ and $c = 2$.
The discriminant is given by:
$ \Rightarrow {b^2} - 4ac = {\left( 6 \right)^2} - 4\left( 7 \right)\left( 2 \right)$
$ \Rightarrow {b^2} - 4ac = 36 - 56$
So the value of the discriminant of the equation is negative,
$\therefore {b^2} - 4ac = - 20$
Here ${b^2} - 4ac < 0$
So the roots of the equation are:
$ \Rightarrow x = \dfrac{{ - 6 \pm \sqrt {{{\left( 6 \right)}^2} - 4\left( 7 \right)\left( 2 \right)} }}{{2\left( 7 \right)}}$
$ \Rightarrow x = \dfrac{{ - 6 \pm \sqrt { - 20} }}{{14}}$
We know that the value of the square root of -1 is equal to $i$, that is $\sqrt { - 1} = i$
So the roots of the equation become as shown below:
$ \Rightarrow x = \dfrac{{ - 6 \pm \sqrt {20\left( { - 1} \right)} }}{{14}}$
Here the number under the root is $\sqrt { - 20} $, it can be written as $\sqrt {20} \sqrt { - 1} = 20i$
$ \Rightarrow x = \dfrac{{ - 6 \pm i2\sqrt 5 }}{{14}}$
So the roots of the equation $7{x^2} + 6x + 2 = 0$ are equal to
$ \Rightarrow x = \dfrac{{ - 6 + i 2\sqrt 5 }}{{14}}$ and $x = \dfrac{{ - 6 - i 2\sqrt 5 }}{{14}}$
$\therefore x = \dfrac{{ - 3 + i\sqrt 5 }}{7}$ and $x = \dfrac{{ - 3 - i\sqrt 5 }}{7}$

Hence the roots of the equation have two imaginary roots.

Note: Please note that in any quadratic equation $a{x^2} + bx + c = 0$, here in this ${b^2} - 4ac$ is called as the determinant. There are 3 general cases of the discriminant.
If ${b^2} - 4ac > 0$, then the quadratic equation has real and distinct roots.
If ${b^2} - 4ac = 0$, then the quadratic equation has real and equal roots.
If ${b^2} - 4ac < 0$, then the quadratic equation has no real roots but has complex conjugate roots.