Question

How do you solve $4{x^2} + 9x + 5 = 0$ using the quadratic formula?

Answer 1

Hint: This is a two degree one variable equation, or simply a quadratic equation. It can be solved with the help of a quadratic formula for that first find the discriminant of the given quadratic equation and then put the values directly in the roots formula of quadratic equation to get the required solution. You will get two solutions for a quadratic equation.

Formula Used:
For a quadratic equation $a{x^2} + bx + c = 0$, discriminant is given as follows
$D = {b^2} - 4ac$
And its roots are given as
$x = \dfrac{{ - b \pm \sqrt D }}{{2a}}$
Use this information to solve the question.$x = \dfrac{{ - b \pm \sqrt D }}{{2a}}$

Complete step by step solution:
To solve the equation $4{x^2} + 9x + 5 = 0$, using quadratic formula we will first find the discriminant of the given equation
Discriminant of a quadratic equation $a{x^2} + bx + c = 0$, is calculated as follows
$D = {b^2} - 4ac$
And roots are given as

For the equation $4{x^2} + 9x + 5 = 0$ values of $a,\;b\;{\text{and}}\;c$ are equals to $4,\;9\;{\text{and}}\;5$ respectively,
$\therefore $ The discriminant for given equation will be
$
 \Rightarrow D = {9^2} - 4 \times 4 \times 5 \\
 \Rightarrow 81 - 80 \\
   \Rightarrow 1 \\
 $
Now finding the solution, by putting all respective values in the root formula
\[
 \Rightarrow x = \dfrac{{ - b \pm \sqrt D }}{{2a}} \\
  \Rightarrow \dfrac{{ - 9 \pm \sqrt 1 }}{{2 \times 4}} \\
  \Rightarrow \dfrac{{ - 9 \pm 1}}{8} \\
 \]
Here we are getting two cases positive and negative, calculating both cases separately
\[
 \Rightarrow x = \dfrac{{ - 9 \pm 1}}{8} \\
   \Rightarrow \dfrac{{ - 9 + 1}}{8}\;{\text{and}}\;\dfrac{{ - 9 - 1}}{8} \\
    \Rightarrow \dfrac{{ - 8}}{8}\;{\text{and}}\;\dfrac{{ - 10}}{8} \\
   \Rightarrow - 1\;{\text{and}}\;\dfrac{{ - 5}}{4} \\
 \]
So the required roots will be given as
$x = - 1\;{\text{and}}\;x = \dfrac{{ - 5}}{4}$

Note: Discriminant of a quadratic equation gives us the information about nature of roots of the equation, if it is greater than zero or positive then roots will be real and distinct (as we have seen in this question), else if it is equals to zero then roots will be real and equal and if it is less than zero or negative then roots will be imaginary.