Question

Find and correct errors of the following mathematical expression:
${\left( {y - 3} \right)^2} = {y^2} - 9$

Answer 1

Hint: In the given problem, to find errors first we will simplify the LHS term and compare it with the RHS term. For this, we will use the expansion of ${\left( {a - b} \right)^2}$ which is given by ${\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2}$.

Complete step by step answer:
In this problem, the mathematical expression ${\left( {y - 3} \right)^2} = {y^2} - 9 \cdots \cdots \left( 1 \right)$ is given.
We have to find and correct the errors in this expression. For this, first we will simplify the LHS term. The LHS term of the equation $\left( 1 \right)$ is ${\left( {y - 3} \right)^2}$.
Now we will use the expansion of ${\left( {a - b} \right)^2}$ which is given by ${\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2}$.
So, by using this formula we can write ${\left( {y - 3} \right)^2} = {\left( y \right)^2} - 2\left( y \right)\left( 3 \right) + {\left( 3 \right)^2}$. Let us simplify this equation.
Therefore, we can write ${\left( {y - 3} \right)^2} = {y^2} - 6y + 9$.
From the equation $\left( 1 \right)$, we can say that the RHS term is ${y^2} - 9$. Now we have the LHS term is ${y^2} - 6y + 9$ and the RHS term is ${y^2} - 9$. So, we can say that LHS is not equal to RHS. If the RHS term of given expression is ${y^2} - 6y + 9$ then we can say that given expression is true.

Therefore, the correct mathematical expression is ${\left( {y - 3} \right)^2} = {y^2} - 6y + 9$.

Note: Expansion of ${\left( {a + b} \right)^2}$ is given by ${\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}$. Expansion of $\left( {a + b} \right)\left( {a - b} \right)$ is given by $\left( {a + b} \right)\left( {a - b} \right) = {a^2} - {b^2}$.