Question

What does this sign mean? $\left( \pm \right)$

Answer 1

Hint: The sign $\left( \pm \right)$ given in the above question is related to the quadratic equations where its solution takes both the negative and the positive values of a number. For this to happen, the quadratic equation needs to be of the type $a{{x}^{2}}+c=0$. On comparing this equation with the standard form of the quadratic equation, which is given by $a{{x}^{2}}+bx+c=0$, we will get $b=0$ as the necessary condition for the solution to be expressed using the given sign $\left( \pm \right)$.

Complete step by step solution:
The sign given in the above question $\left( \pm \right)$ indicates the combination of the positive and the negative sign. It is used to represent the value of a variable which takes both the positive and the negative values of a number. The value of a variable occurs in this form when the variable is represented by the quadratic equation of the form
$\Rightarrow a{{x}^{2}}+c=0$
On comparing the above equation with the standard form of a quadratic equation which is given by $a{{x}^{2}}+bx+c=0$, we get $b=0$. This means that the necessary and sufficient condition for the value of a variable to be expressed using the $\left( \pm \right)$ sign is that it must be expressed in the form of a quadratic equation in which coefficient of x must be equal to zero.
Let us consider an example of this type of equation to understand it better. Let us choose $a=4$ and $c=-9$ to write the below equation as
$\Rightarrow 4{{x}^{2}}-9=0$
Adding $9$ on both the sides of the above equation we get
\[\begin{align}
  & \Rightarrow 4{{x}^{2}}-9+9=0+9 \\
 & \Rightarrow 4{{x}^{2}}=9 \\
\end{align}\]
Now, dividing the both sides of the above equation by \[4\] we get
$\begin{align}
  & \Rightarrow \dfrac{4{{x}^{2}}}{4}=\dfrac{9}{4} \\
 & \Rightarrow {{x}^{2}}=\dfrac{9}{4} \\
\end{align}$
Now, we subtract $\dfrac{9}{4}$ from both the sides to get
$\begin{align}
  & \Rightarrow {{x}^{2}}-\dfrac{9}{4}=\dfrac{9}{4}-\dfrac{9}{4} \\
 & \Rightarrow {{x}^{2}}-\dfrac{9}{4}=0 \\
\end{align}$
Writing $\dfrac{9}{4}={{\left( \dfrac{3}{2} \right)}^{2}}$ we get
$\Rightarrow {{x}^{2}}-{{\left( \dfrac{3}{2} \right)}^{2}}=0$
Now, by using the algebraic identity ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$ we can write the above equation as
\[\Rightarrow \left( x+\dfrac{3}{2} \right)\left( x-\dfrac{3}{2} \right)=0\]
Finally, using the zero product rule we can write
$\Rightarrow x=-\dfrac{3}{2}$, and
$\Rightarrow x=\dfrac{3}{2}$
Since x takes both negative and the positive values, we can combine these two values two write
$\Rightarrow x=\pm \dfrac{3}{2}$
Hence, we understood the meaning of the $\left( \pm \right)$ sign.

Note: It is a common misconception to consider the square root of a number to be expressed using the $\left( \pm \right)$ sign. For example, one may write $\sqrt{4}=\pm 2$. But this equation is incorrect. This is because the square root of a number is always a non-negative number. The use of the $\left( \pm \right)$ sign is only there in the quadratic equations of the type shown in the above solution.