Question

Solve the particular equation $2{x^2} - 8 = 0$

Answer 1

Hint: In order to solve this equation first we have to take common terms so that this is easy to solve and further try to solve it by doing factorization after that we have to find zeros and get the values of Variable in the equation.

Complete step-by-step answer:
Given equation is $2{x^2} - 8 = 0$
In this equation we have to find out the values of $x$
Here in this equation we can take $2$ as common so equation becomes
$
  2{x^2} - 8 = 0 \\
   \Rightarrow 2({x^2} - 4) = 0 \\
$
Further solving this
\[
  2({x^2} - 4) = 0 \\
   \Rightarrow ({x^2} - 4) = \dfrac{0}{4} \\
\]
We know that $\dfrac{0}{{{\text{constant}}}}$ always becomes $0$
So we can write this equation as
$
   \Rightarrow ({x^2} - 4) = \dfrac{0}{4} \\
  \because \dfrac{0}{4} = 0 \\
   \Rightarrow ({x^2} - 4) = 0 \\
$
Further solving this equation
$\because ({a^2} - {b^2}) = (a + b)(a - b)$
$
  ({x^2} - 4) = 0 \\
   \Rightarrow ({x^2} - {(2)^2}) = 0 \\
   \Rightarrow (x + 2)(x - 2) = 0 \\
$
${\text{Either }}(x + 2) = 0{\text{ or }}(x - 2) = 0$
If $(x + 2) = 0$
$
  (x + 2) = 0 \\
   \Rightarrow x = - 2 \\
$
If $(x - 2) = 0$
$ (x - 2) = 0 \\
   \Rightarrow x = 2 \\
$
So we get $x = 2$ and $x = - 2$.
Hence, $x = 2, - 2$

Note- In algebra, a quadratic equation is any equation that can be rearranged in standard form as $a{x^2} + bx + c = 0$ $a{x^2} + bx + c = 0$ where $x$ represents an unknown, here $a,b,$ and $c$ represent known numbers, where \[a \ne 0\]. If \[a = 0,\] then the equation is linear, not quadratic, as there is no term.