Question

How do you solve $6{x^2} - 7x - 3 = 0$ using the quadratic formula?

Answer 1

Hint:
Given an expression. We have to find the solution to the equation. First, we will rewrite the equation in the standard form of quadratic equation. We will apply the quadratic formula to the equation. Then, we will find the value of x.

Formula used:
The quadratic formula to find the value of x in the quadratic equation of the form $a{x^2} + bx + c = 0$ is given by:
$x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$

Complete step by step solution:
We are given the polynomial $6{x^2} - 7x - 3 = 0$.
Now, we will factorize the equation by applying the quadratic formula because splitting the middle term to factorize the equation does not work in this equation.
Compare the equation with standard form of equation, $a{x^2} + bx + c = 0$ we get:
Substitute $a = 6$, $b = - 7$ and $c = - 3$ into the formula, we get:
$x = \dfrac{{ - \left( { - 7} \right) \pm \sqrt {{{\left( { - 7} \right)}^2} - 4\left( 6 \right)\left( { - 3} \right)} }}{{2\left( 6 \right)}}$
On simplifying the terms of the expression, we get:
$ \Rightarrow x = \dfrac{{7 \pm \sqrt {49 + 72} }}{{12}}$
$ \Rightarrow x = \dfrac{{7 \pm \sqrt {121} }}{{12}}$
Further simplify the expression, we get:
$ \Rightarrow x = \dfrac{{7 \pm 11}}{{12}}$
Compute the roots of the equation.
$ \Rightarrow x = \dfrac{{7 + 11}}{{12}}{\text{ or }}x = \dfrac{{7 - 11}}{{12}}$
$ \Rightarrow x = \dfrac{{18}}{{12}}{\text{ or }}x = \dfrac{{ - 4}}{{12}}$
Cancel out the common terms, we get:
$ \Rightarrow x = \dfrac{3}{2}{\text{ or }}x = - \dfrac{1}{3}$

Final answer: Hence the solutions of the equation are $x = - \dfrac{1}{3}$ and $x = \dfrac{3}{2}$

Additional Information:
In the method of splitting the middle term, the sum of two terms must be equal to middle terms and the product of two terms is equal to the last term. This method is only employed when the middle terms can be easily broken down into two terms.

Note:
In such types of questions students mainly make mistakes while factorization of the equation. Students may get confused to apply the method which is appropriate to find the factors of the equation. Such types of questions can be solved using various methods such as splitting of the middle term.