Question

How do you solve $2{x^2} + 9x + 4 = 0$ using the quadratic formula?

Answer 1

Hint: In this question, we are given a polynomial equation and we have to solve it, that is, we have to find its factors. The degree of a polynomial equation is defined as the highest exponent of the unknown quantity in a polynomial equation. The factors/solution/zeros of the given polynomial are defined as those values of x at which the value of a polynomial is zero. The equation given in the question has a degree 2, so we have a quadratic equation and we are already told to solve the given quadratic equation using the quadratic formula, so for that, we will first express the given equation in the standard equation form and then the values of the coefficients are plugged in the quadratic formula.

Complete step by step solution:
The equation given is $2{x^2} + 9x + 4 = 0$
On comparing the given equation with the standard quadratic equation $a{x^2} + bx + c = 0$ , we get –
$a = 2,\,b = 9,\,c = 4$
The Quadratic formula is given as –
$x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
Now, put the known values in the above equation –
$
  x = \dfrac{{ - 9 \pm \sqrt {{{(9)}^2} - 4(2)(4)} }}{{2(2)}} \\
   \Rightarrow x = \dfrac{{ - 9 \pm \sqrt {81 - 32} }}{4} \\
   \Rightarrow x = \dfrac{{ - 9 \pm \sqrt {49} }}{4} \\
   \Rightarrow x = \dfrac{{ - 9 \pm 7}}{4} \\
   \Rightarrow x = \dfrac{{ - 9 + 7}}{4},\,x = \dfrac{{ - 9 - 7}}{4} \\
   \Rightarrow x = - \dfrac{1}{2},\,x = - 4 \\
 $

Hence the zeros of the given equation are $ - \dfrac{1}{2}$ and $ - 4$ .

Note: When the alphabets representing an unknown variable quantity in an algebraic expression are raised to some non-negative integer as the power, the algebraic expression becomes a polynomial equation. A quadratic polynomial is defined as a polynomial of degree two and its zeros can be found using many methods like factorization, completing the square, graphs, quadratic formula etc. We use the quadratic formula when we fail to find the factors of the equation. But in this question we can also solve the equation by factorization as the factors of the equation can be found easily by hit and trial.