Question

How do you solve \[4\left| 0.5x-2.5 \right|=0\]?

Answer 1

Hint: Take 0.5 common from the terms inside the modulus sign and take it out of the modulus. Multiply this 0.5 with 4 to simplify the equation. Divide both the sides with the obtained product. Now, use the property of modulus given as: - if \[\left| x \right|=a\] then \[x=\pm a\], to remove the modulus sign. Solve the obtained linear equation to get the value of x.

Complete step-by-step answer:
Here, we have been provided with the equation \[4\left| 0.5x-2.5 \right|=0\] and we are asked to solve it. That means we have to find the value of x.
As we can see that the given equation is a linear equation in one variable, that is x. Also, this equation contains the modulus sign. So, to solve this equation, first we need to remove the modulus sign. So, we have,
\[\Rightarrow 4\left| 0.5x-2.5 \right|=0\]
Taking 0.5 common from the terms inside the modulus sign, we get,
\[\Rightarrow 4\left| 0.5\left( x-5 \right) \right|=0\]
Here, we can take 0.5 out of the modulus sign, so we get,
\[\begin{align}
  & \Rightarrow 4\times 0.5\left| \left( x-5 \right) \right|=0 \\
 & \Rightarrow 2.0\left| \left( x-5 \right) \right|=0 \\
 & \Rightarrow 2\left| x-5 \right|=0 \\
\end{align}\]
Dividing both the sides with 2 and using the property that ‘any non – zero real number when divides 0 gives 0’ we get,
\[\begin{align}
  & \Rightarrow \dfrac{2\left| x-5 \right|}{2}=\dfrac{0}{2} \\
 & \Rightarrow \left| x-5 \right|=0 \\
\end{align}\]
Now, we know that if \[\left| x \right|=a\] then \[x=\pm a\]. Here, ‘a’ must be a positive number and it may be 0 also. We know that modulus of any number, whether positive or negative, always gives a positive form of that number. This is the reason that ‘a’ cannot be negative. So, applying the above stated property we get,
\[\begin{align}
  & \Rightarrow x-5=\pm 0 \\
 & \Rightarrow x-5=0 \\
\end{align}\]
\[\Rightarrow x=5\]
Hence, the value of x is 5.

Note: One may note that we have obtained only a single value of x. This is because in the R.H.S. We were provided with 0. If it were any positive real number then we would have obtained two values of x. Note that we also solve many inequalities consisting of modulus signs. In such cases, first we substitute the expression inside the modulus sign equal to 0 and then solve for the value of x which is then represented on the number line. Now, we consider different cases and solve the inequality. You must remember all the properties of modulus because equations are easy to solve but inequations require the use of certain properties. Always remember that if ‘a’ is negative then \[\left| a \right|=-a\]. For example: - \[\left| -1 \right|=-\left( -1 \right)=+1\].