Question

How do you solve $6{x^2} + 7x + 2 = 0$ ?

Answer 1

Hint: To solve these questions, first write the given equation in the standard form of quadratic equations, that is, $a{x^2} + bx + c = 0$ where $a \ne 0$ . In this form $a$ and $b$ are the coefficients of the terms ${x^2}$ and $x$ respectively and $c$ is the constant term.

Formula used: The roots of a quadratic equation $a{x^2} + bx + c = 0$ can be found by using the given formula,
$x = \left[ {\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}} \right]$

Complete step-by-step solution:
The given expression is $6{x^2} + 7x + 2 = 0$.
Now, compare it with the standard form $a{x^2} + bx + c = 0$ to get,
$a = 6$ , $b = 7$ and $c = 2$ .
The roots of the quadratic equation ( $a{x^2} + bx + c = 0$ ) can be found by the formula,
$x = \left[ {\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}} \right]$
On substituting the values of $a$ , $b$ and $c$ in the Quadratic formula, we get
$\Rightarrow x = \left[ {\dfrac{{ - 7 \pm \sqrt {{7^2} - 4 \times 6 \times 2} }}{{2 \times 6}}} \right]$
Evaluate the expression inside the root first.
$\Rightarrow x = \left[ {\dfrac{{ - 7 \pm \sqrt {49 - 48} }}{{12}}} \right]$
On simplifying we get,
$\Rightarrow x = \left[ {\dfrac{{ - 7 \pm \sqrt 1 }}{{12}}} \right]$
Solve the root part entirely and perform further simplification,
$\Rightarrow x = \left[ {\dfrac{{ - 7 \pm 1}}{{12}}} \right]$
This now simplifies to two cases,
$\Rightarrow x\left[ {\dfrac{{ - 7 + 1}}{{12}}} \right]$ and $\Rightarrow x = \left[ {\dfrac{{ - 7 - 1}}{{12}}} \right]$
Evaluate further to get simple fractions,
$\Rightarrow x = \dfrac{{ - 6}}{{12}}$ and $\Rightarrow x = \dfrac{{ - 8}}{{12}}$
$\Rightarrow x = - \dfrac{1}{2}$ and $\Rightarrow x = - \dfrac{2}{3}$

Therefore, the solution of the equation $6{x^2} + 7x + 2 = 0$ is $x = - \dfrac{1}{2}, - \dfrac{2}{3}$

Note: These questions can also be solved by one more method, known as splitting the middle term. . In this method the middle term, which is the $x$ term is split into two factors and product equal to the last term. This method is useful only when the middle term can be split into the sum of the product of the first and last term.
Always remember to take two conditions whenever there is a $\pm$ symbol. The expression is written twice once with $\;+$ and another time with $\;-$ . You can always cross-check your answer by placing the value of $x$ back in the equation. If you get $L.H.S. = R.H.S.$ , then your answer is correct.