Question

How do you solve $ 3{x^2} + 5x - 2 = 0 $ using the quadratic formula?

Answer 1

Hint: First we will reduce the equation further if possible. Then we will try to factorise the terms in the equation. Then we will use the quadratic formula to solve for the value of $ x $ using the formula which is given by $ x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} $ .

Complete step-by-step answer:
We will start off by taking all the terms to one side.
 $ 3{x^2} + 5x - 2 = 0 $
Now we will use the quadratic formula $ x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} $ .
But first we identify the values of the terms used in the formula by comparing the values with the general form of the quadratic equation which is given by $ a{x^2} + bx + c = 0 $ .
After comparing we get the values as:
  $
  a = 3 \\
  b = 5 \\
  c = - 2 \;
  $
Now we substitute the terms in the formula.
 $
  x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} \\
  x = \dfrac{{ - (5) \pm \sqrt {{{(5)}^2} - 4(3)( - 2)} }}{{2(3)}} \\
  x = \dfrac{{ - 5 \pm \sqrt {{{(5)}^2} + 4(3)(2)} }}{6} \;
  $
Now we simplify further for the value of $ x $ .
 $
  x = \dfrac{{ - 5 \pm \sqrt {{{(5)}^2} + 4(3)(2)} }}{6} \\
  x = \dfrac{{ - 5 \pm \sqrt {25 + 24} }}{6} \\
  x = \dfrac{{ - 5 \pm \sqrt {49} }}{2} \\
  x = \dfrac{{ - 5 \pm 7}}{2} \;
  $
Now to solve for the unknown variable, we separate the equations into two equations; one with a plus sign and the other with the negative sign and solve for the value of $ x $ .

\[x = \dfrac{{ - 5 - 7}}{6} = \dfrac{{ - 12}}{6} = - 2\]
\[x = \dfrac{{ - 5 + 7}}{6} = \dfrac{2}{6} = \dfrac{1}{3}\]
Hence, the values of the unknown variable $ x $ are $ - 2,\dfrac{1}{3} $ .
So, the correct answer is “ $ - 2,\dfrac{1}{3} $ ”.

Note: While comparing the values of the given equation with the general form of quadratic equation which is given by $ a{x^2} + bx + c = 0 $ , compare along with their respective signs. While applying the quadratic formula, make sure you substitute all the values along with their respective signs. Solve all the equations separately, so that you don’t miss any term of the solution. Check if the solution satisfies the original equation completely. If any term of the solution doesn’t satisfy the equation, then that term will not be considered as a part of solution