Question

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

Answer 1

Hint: In this question we will compare the given equation with the general quadratic equation. Then we have to use the quadratic formula to find the required roots of the given quadratic equation. Finally we get the required answer.

Formula used: $({x_1},{x_2}) = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$, where ${x_1}$ and ${x_2}$ are two roots of the equation.

Complete step-by-step solution:
We have the given quadratic equation as:
$ \Rightarrow 2{x^2} + x - 1 = 0$
On comparing this quadratic equation with the general form of a quadratic equation $a{x^2} + bx + c = 0$,
We get the values of $a,b$ and $c$ as:
$a = 2$,$b = 1$ and $c = - 1$.
Now on substituting the values in the quadratic formula, we get:
$ \Rightarrow ({x_1},{x_2}) = \dfrac{{ - 1 \pm \sqrt {{1^2} - 4 \times 2 \times - 1} }}{{2 \times 2}}$
on simplifying the root part and the denominator of the equation, we get:
\[ \Rightarrow ({x_1},{x_2}) = \dfrac{{ - 1 \pm \sqrt {1 + 8} }}{4}\]
on simplifying we get:
\[ \Rightarrow ({x_1},{x_2}) = \dfrac{{ - 1 \pm \sqrt 9 }}{4}\]
Now we know that the value of $\sqrt 9 = 3$, therefore we can write the term as:
\[ \Rightarrow ({x_1},{x_2}) = \dfrac{{ - 1 \pm 3}}{4}\]
Now on splitting the equation, we get:
\[ \Rightarrow {x_1} = \dfrac{{ - 1 + 3}}{4}\] and \[{x_2} = \dfrac{{ - 1 - 3}}{4}\]
Now on simplifying we get:
\[ \Rightarrow {x_1} = \dfrac{2}{4}\] and \[{x_2} = \dfrac{{ - 4}}{4}\]

Therefore, ${x_1} = \dfrac{1}{2}$ and ${x_2} = - 1$, are the two roots of the equation.

Note: Now to check whether the roots we have found are correct, we will substitute the values of the roots in the right-hand side of the quadratic equation to verify it:
On substituting $x = \dfrac{1}{2}$ in the left-hand side of the quadratic equation, we get:
$ \Rightarrow 2{\left( {\dfrac{1}{2}} \right)^2} + \dfrac{1}{2} - 1$
On simplifying, we get:
$ \Rightarrow \dfrac{2}{4} + \dfrac{1}{2} - 1 = 0$ therefore, this root is correct.
Now on substituting $x = - 1$ in the left-hand side of the quadratic equation, we get:
$ \Rightarrow 2{\left( { - 1} \right)^2} - 1 - 1$
On simplifying, we get:
$ \Rightarrow 2 - 1 - 1 = 0$ therefore, this root is also correct.
It is to be remembered that all the quadratic equations do not have real roots, if there is a negative number generated in the root section of the quadratic formula which is $\sqrt {{b^2} - 4ac} $, then the solution would become a complex number which has the imaginary part $i$ present in it.