Question

How do you solve $\left| 3z-4 \right|=\left| 5z-6 \right|$?

Answer 1

Hint: For solving the given equation, we first need to remove the modulus sign by taking the squares of both the sides. Then we have to use the algebraic identity ${{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}$ to expand both the sides. Then on simplifying the obtained equation by using the algebraic manipulations, we will obtain a quadratic equation. Using the middle term splitting method, we can solve the quadratic equation and hence obtain the solutions of the given equation.

Complete step-by-step solution:
The given equation is written as
$\Rightarrow \left| 3z-4 \right|=\left| 5z-6 \right|$
Taking squares of both sides of the above equation, we get
$\Rightarrow {{\left( 3z-4 \right)}^{2}}={{\left( 5z-6 \right)}^{2}}$
Now, we know the algebraic identity ${{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}$. Using this identity, the above equation can also be written as
$\begin{align}
  & \Rightarrow {{\left( 3z \right)}^{2}}-2\left( 3z \right)\left( 4 \right)+{{\left( 4 \right)}^{2}}={{\left( 5z \right)}^{2}}-2\left( 5z \right)\left( 6 \right)+{{\left( 6 \right)}^{2}} \\
 & \Rightarrow 9{{z}^{2}}-24z+16=25{{z}^{2}}-60z+36 \\
\end{align}$
Subtracting $9{{z}^{2}}-24z+16$ from both the sides, we get
\[\begin{align}
  & \Rightarrow 9{{z}^{2}}-24z+16-\left( 9{{z}^{2}}-24z+16 \right)=25{{z}^{2}}-60z+36-\left( 9{{z}^{2}}-24z+16 \right) \\
 & \Rightarrow 0=25{{z}^{2}}-60z+36-9{{z}^{2}}+24z-16 \\
 & \Rightarrow 0=16{{z}^{2}}-36z+20 \\
 & \Rightarrow 16{{z}^{2}}-36z+20=0 \\
\end{align}\]
Dividing both the sides of the above equation by \[4\] we get
$\Rightarrow 4{{z}^{2}}-9z+5=0$
The above equation is a quadratic equation in $z$. Splitting the middle term as $-9z=-4z-5z$ we get
\[\Rightarrow 4{{z}^{2}}-4z-5z+5=0\]
Taking \[4z\] common from the first two terms and $-5$ common from the last two terms, we get
$\Rightarrow 4z\left( z-1 \right)-5\left( z-1 \right)=0$
Now taking $\left( z-1 \right)$ common, we get
$\begin{align}
  & \Rightarrow \left( z-1 \right)\left( 4z-5 \right)=0 \\
 & \Rightarrow \left( z-1 \right)=0,\left( 4z-5 \right)=0 \\
\end{align}$
On solving, we finally get
$\Rightarrow z=1,z=\dfrac{5}{4}$
Hence, the solutions of the given equation are $z=1$ and $z=2$.

Note: By observing the variable $z$ in the given equation, we may think that it is a complex variable. But we must not consider it to be a complex variable unless stated in the question. We can also solve the quadratic equation obtained above by using the quadratic formula $z=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ instead of the middle term splitting method. Also, do not forget to check the final solutions by substituting them back into the original given equation.