Question

How do you solve $4{x^2} - 6x - 10 = 0$ using the quadratic formula?

Answer 1

Hint: In this given problem to solve the equation and find the value of $x$ using quadratic formula, first simplify the equation such that all terms should be on left hand side and then compare the quadratic coefficients with coefficients of the given equation and then simply put the values in the quadratic formula to get the solution.
Quadratic formula for the equation $a{x^2} + bx + c = 0$ is given as follows
$x = \dfrac{{ - b \pm \sqrt D }}{{2a}}$
Where $D$ is the discriminant of the quadratic equation and calculated as $D = \sqrt {{b^2} - 4ac} $

Complete step by step solution:
In order to solve the given expression $4{x^2} - 6x - 10 = 0$ with help of quadratic formula, we will first compare the coefficients of the given quadratic with standard quadratic equation to get the value of quadratic coefficients as follows
\[4{x^2} - 6x - 10 = 0\;{\text{and}}\;a{x^2} + bx + c = 0\]
Therefore respective values of the quadratic coefficients are
$a = 4,\;b = - 6\;{\text{and}}\;c = - 10$
Now we know that the solution for the quadratic equation $a{x^2} + bx + c = 0$ is given as
$x = \dfrac{{ - b \pm \sqrt D }}{{2a}}$
Where $D$ is the discriminant of the quadratic expression, which can be calculated as
$D = \sqrt {{b^2} - 4ac} $
So first finding the value of discriminant,
$
  D = {( - 6)^2} - 4 \times 4 \times ( - 10) \\
  D = 36 + 160 \\
  D = 196 \\
 $
Now substituting all the respective values in quadratic formula in order to get the solution for $x$
$
  x = \dfrac{{ - ( - 6) \pm \sqrt {196} }}{{2 \times 4}} \\
   = \dfrac{{9 \pm 14}}{8} \\
   = \dfrac{{6 + 14}}{8}\;{\text{and}}\;\dfrac{{6 - 14}}{8} \\
   = \dfrac{{20}}{8}\;{\text{and}}\;\dfrac{{ - 8}}{8} \\
   = \dfrac{5}{2}\;{\text{and}}\; - 1 \\
 $
Therefore $x = \dfrac{5}{2}\;{\text{and}}\;x = - 1$ are the required solutions for the equation $4{x^2} - 6x - 10 = 0$


Note: Discriminant of a quadratic equation gives us an idea about the nature of roots the quadratic equation will have. If discriminant is negative then roots will be imaginary, if discriminant equals zero then roots will be real and equal, and if positive then roots will be real and distinct as in this case.