Question

How do you solve $2{x^2} - 3x - 1 = 0$ using the quadratic formula?

Answer 1

Hint: First of all, we will compare the given equation with the standard quadratic equation, i.e., $a{x^2} + bx + c = 0;a \ne 0$ and after comparing both the equations with each other, we will find the values of $a$ , $b$ and $c$ . Then, we will be substituting the values in the quadratic formula which is mentioned below.

Formula used:
Quadratic formula: For $a{x^2} + bx + c = 0;a \ne 0$ , the values of $x$ which are the solutions of the equation are given by: $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$ .

Complete step-by-step answer:
First of all, we need to solve the given quadratic equation. Now, in order to solve the given quadratic equation, we first need to compare the given quadratic equation with the standard quadratic equation.
Now, the standard quadratics equation is $a{x^2} + bx + c = 0$ .
Let us suppose, $2{x^2} - 3x - 1 = 0$ - - - - - - - - - - $(1.)$
Now, let’s compare the standard quadratic equation with the given quadratic equation, which is represented as $(1.)$
After comparing both the equations with each other, we get,
$a = 2$ , $b = - 3$ and $c = - 1$ .
As we need to find the values of $x$ , let us substitute these values in the quadratic formula, which is $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
After substituting the values in the above formula, it becomes
$ \Rightarrow \dfrac{{ - ( - 3) \pm \sqrt {{{\left( { - 3} \right)}^2} - 4(2)( - 1)} }}{{2(2)}}$
$ \Rightarrow \dfrac{{3 \pm \sqrt {9 + 8} }}{4}$
Adding numbers inside the square root, we get
$ \Rightarrow \dfrac{{3 \pm \sqrt {17} }}{4}$
∴ $x = \dfrac{{3 \pm \sqrt {17} }}{4}$
Thus we can say that, $x = \dfrac{{3 + \sqrt {17} }}{4}$ or $x = \dfrac{{3 - \sqrt {17} }}{4}$ .

Hence, the quadratics equation $2{x^2} - 3x - 1 = 0$ after solving with quadratic formula has the roots $x = \dfrac{{3 + \sqrt {17} }}{4}$ or $x = \dfrac{{3 - \sqrt {17} }}{4}$.
Additional Information: Depending upon the quadratic equation, the nature of roots varies.
If the discriminant of the quadratic equation is positive or greater than zero, then the two roots of any quadratic equation are real.
If the discriminant of the quadratic equation is zero, then the roots of the quadratic equation are real.
If the discriminant of the quadratic equation is negative or less than zero, then the quadratic equation has no real roots.

Note:
In the standard form of a quadratic equation which is, $a{x^2} + bx + c = 0,a \ne 0$ , then the expression ${b^2} - 4ac$ is known as its discriminant and is generally denoted by $D$ . Now, depending upon the quadratic equation, the value of the discriminant varies and hence the nature of roots also varies.