Question

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

Answer 1

Hint: In this given problem to solve the equation and find the value of $x$. Use the quadratic formula in order to solve this equation, first compare the quadratic coefficients with the coefficients of the given equation and then simply put the values in the quadratic formula to get the solution.

Formula used: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 ${x^2} - 3 = 0$ with help of quadratic formula, we need to first find the values of respective quadratic coefficients of the given expression.
To find quadratic coefficients of the expression comparing it with the standard quadratic equation
${x^2} - 3 = 0\;{\text{and}}\;a{x^2} + bx + c$
We can also write it as
${x^2} + 0x + ( - 3) = 0\;{\text{and}}\;a{x^2} + bx + c$
Therefore respective values of quadratic coefficients are
$a = 1,\;b = 0\;{\text{and}}\;c = - 3$
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 = {0^2} - 4 \times 1 \times ( - 3) \\
  D = 12 \\
 $
Now substituting all the respective values in quadratic formula in order to get the solution for $x$
$
  x = \dfrac{{ - 0 \pm \sqrt {12} }}{{2 \times 1}} \\
   = \dfrac{{ \pm 2\sqrt 3 }}{2} \\
   = \pm \sqrt 3 \\
 $
Therefore $x = \sqrt 3 \;{\text{and}}\;x = - \sqrt 3 $ are the required solutions for the equation ${x^2} - 3 = 0$.

Note: Quadratic formula is only applicable for quadratic equations that are equations with degree two. This problem can be solved by two more methods, one with the help of algebraic identity of difference between two square numbers and the other is algebraically by directly taking the root of both sides after sending constant terms to the opposite side to the variable.