Question

How do you solve ${{\left( \sqrt{y}-5 \right)}^{2}}$ ?

Answer 1

Hint: As the given expression is of the form ${{\left( a-b \right)}^{2}}$, so we will use the algebraic formula ${{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab$ to solve the given expression. By substituting the values in the formula and simplifying further we get the desired answer.

Complete step by step solution:
We have been given an expression ${{\left( \sqrt{y}-5 \right)}^{2}}$.
We have to simplify the given expression.
Now, the given expression is of the form ${{\left( a-b \right)}^{2}}$. So by comparing the given expression we get the values
$a=\sqrt{y}$ and $b=5$
Now, we know that the algebraic identity is given as ${{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab$
Now, substituting the values in the formula we will get
$\Rightarrow {{\left( \sqrt{y} \right)}^{2}}+{{5}^{2}}+2\times \sqrt{y}\times 5$
As we know that square and square root cancel each other.
Now, simplifying the above obtained equation we will get
$\Rightarrow y+25+10\sqrt{y}$
Hence above is the required simplified form of the given expression.

Note:
To solve such types of questions one must have the knowledge of algebraic identities and formulas. The possibility of mistake while simplifying the equation is that after applying the formula students may write the equation as
$\begin{align}
  & \Rightarrow {{\left( \sqrt{y} \right)}^{2}}+{{5}^{2}}+2\times \sqrt{y}\times 5 \\
 & \Rightarrow {{y}^{2}}+25+10\sqrt{y} \\
\end{align}$
Students may consider the obtained equation as a quadratic equation and solve further, which is incorrect as the square and square root cancel each other and we get the simplified equation as $y+25+10\sqrt{y}$ which is not a quadratic equation and we cannot solve the equation further.
We can further simplify the obtained equation as:
$\Rightarrow y+10\sqrt{y}=-25$