Question

How do you find the missing number or term to make a perfect square trinomial: \[{{y}^{2}}+\dfrac{1}{2}y+.....\]?

Answer 1

Hint: Convert \[\dfrac{1}{2}y\] into the form 2ab by multiplying and dividing with 2. Now, compare 2 with 2, y with ‘a’ and the obtained fraction with b and use the algebra identity: - \[{{a}^{2}}+{{b}^{2}}+2ab={{\left( a+b \right)}^{2}}\] to make the given trinomial a perfect square.

Complete step by step answer:
Here, we have been provided with the expression: - \[{{y}^{2}}+\dfrac{1}{2}y+.....\] and we have been asked to find the missing number or term so that we can make the trinomial a perfect square.
Now, we know that there are two basic whole square formulas for a binomial expression. They are: - \[{{a}^{2}}+{{b}^{2}}+2ab={{\left( a+b \right)}^{2}}\] and \[{{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab\]. For the given expression \[{{y}^{2}}+\dfrac{1}{2}y+.....\], let us assume a = y. We need to find ‘b’. Now, as we can say that here we are going to use the formula for \[{{\left( a+b \right)}^{2}}\] because here we have the term +2ab which is similar to \[+\dfrac{1}{2}y\], both have positive signs, so we have,
\[\Rightarrow {{y}^{2}}+\dfrac{1}{2}y+.....\]
We can write it as, by multiplying the second term with 2 and to balance it dividing it with 2,
\[\begin{align}
  & \Rightarrow {{y}^{2}}+2\times \dfrac{1}{2}y\times \dfrac{1}{2}+..... \\
 & \Rightarrow {{y}^{2}}+2\times y\times \dfrac{1}{4}+..... \\
\end{align}\]
So, on comparing it with \[{{a}^{2}}+{{b}^{2}}+2ab\], we can conclude that,
\[\Rightarrow a=y\] and \[b=\dfrac{1}{4}\]
That means we have to add the term \[{{b}^{2}}=\dfrac{1}{{{\left( 4 \right)}^{2}}}\] to make it a perfect square. So, adding \[\dfrac{1}{{{\left( 4 \right)}^{2}}}\], we get,
\[\Rightarrow {{y}^{2}}+2\times \dfrac{1}{y}\times \dfrac{1}{4}+\dfrac{1}{{{4}^{2}}}={{\left( y+\dfrac{1}{4} \right)}^{2}}\]

Hence, the missing number is \[\dfrac{1}{{{\left( 4 \right)}^{2}}}=\dfrac{1}{16}\] which is our answer.

Note: One must remember the two most important and basic algebraic identities given as: - \[{{a}^{2}}+{{b}^{2}}+2ab={{\left( a+b \right)}^{2}}\] and \[{{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab\]. Remember that the above approach that we have applied to solve the question is known as completing the square method which is a very important part of quadratic equation. The discriminant formula that we used to solve the quadratic equation is derived using the approach of completing the square method. This method is further used in coordinate geometry of parabola for finding the vertex and axis of the parabola.