Question

How do you solve ${{\left( x-2 \right)}^{2}}+95={{\left( x-3 \right)}^{2}}$ ?

Answer 1

Hint: To solve ${{\left( x-2 \right)}^{2}}+95={{\left( x-3 \right)}^{2}}$ , we will use algebraic rules and identities. We know that ${{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}$ . We have to apply this in the given equation and simplify. By solving for x, we will get the required solution.

Complete step by step solution:
We need to solve ${{\left( x-2 \right)}^{2}}+95={{\left( x-3 \right)}^{2}}$ . For this, we will use algebraic rules and identities.
We have to first simplify the square terms. We know that ${{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}$ . hence, we can write the given equation as
$\begin{align}
  & {{\left( x-2 \right)}^{2}}+95={{\left( x-3 \right)}^{2}} \\
 & \Rightarrow {{x}^{2}}-2\times 2\times x+{{2}^{2}}+95={{x}^{2}}-2\times 3\times x+{{3}^{2}} \\
\end{align}$
We can simplify this further as shown below.
$\Rightarrow {{x}^{2}}-4x+4+95={{x}^{2}}-6x+9$
We can see that ${{x}^{2}}$ is common on both sides. So, let us cancel it.
$\Rightarrow -4x+4+95=-6x+9$
Now, we have collected x terms on LHS and constants on RHS.
$\Rightarrow -4x+6x=9-4-95$
Let us simplify this. We will get
$\begin{align}
  & \Rightarrow 2x=5-95 \\
 & \Rightarrow 2x=-90 \\
\end{align}$
We can find x by taking 2 from LHS to RHS.
$\Rightarrow x=\dfrac{-90}{2}=-45$
Hence, the value of x is -45.

Note: Students must know algebraic rules and identities to simplify equations. Students have a chance of making mistake by writing the formula for ${{\left( a-b \right)}^{2}}$ as ${{a}^{2}}+2ab-{{b}^{2}}$ or ${{a}^{2}}+2ab+{{b}^{2}}$ . They must be careful when doing the calculations. When we move a positive term LHS to RHS or vice-versa, that term becomes negative. Similarly, when we move a negative term from LHS to RHS or vice-versa, that term becomes positive. When we move a product term from one side to the other, it becomes the divisor. Similarly, when we move a divisor from one side to another, it will be a multiplier.