Question

Find the zeros of the following quadratic polynomial and verify the relationship between the zeros and the coefficients.
\[4{s^2} - 4s + 1\]

Answer 1

Hint:
To solve the above question, first of all, we will find the ‘zeros’ i.e. the roots of the given quadratic equation, using factorization. We will first equate the given expression to 0. Then, we will factorize the equation to find the zeros of the polynomial. Then we will use the formula of sum of roots and product of roots to verify the relationship.

Complete step by step solution:
The given polynomial is:
\[4{s^2} - 4s + 1\]
Now, according to the question, we need to find the zeros of the given polynomial i.e. we have to find the value of the variable \[s\]. So we will equate the given expression to 0.
Therefore, we get
\[ \Rightarrow 4{s^2} - 4s + 1 = 0\]
Now, we can simplify this quadratic polynomial by using the middle term splitting method.
Middle term splitting is a method in which we divide the middle term in two factors such that their product equals to the product of the first and the last term.
\[ \Rightarrow 4{s^2} - 2s - 2s + 1 = 0\]
Factoring out terms, we get
\[ \Rightarrow 2s\left( {2s - 1} \right) - 1\left( {2s - 1} \right) = 0\]
Again factorizing the above equation, we get
\[ \Rightarrow \left( {2s - 1} \right)\left( {2s - 1} \right) = 0\]
\[ \Rightarrow \left( {2s - 1} \right) = 0\] or \[\left( {2s - 1} \right) = 0\]
Adding \[1\] on both sides of the above two equations, we get,
\[ \Rightarrow 2s = 1\] or \[2s = 1\]
Dividing both sides by \[2\],
\[ \Rightarrow s = \dfrac{1}{2}\], \[\dfrac{1}{2}\]
Hence, the zeros of the polynomial \[4{s^2} - 4s + 1 = 0\] are \[s = \dfrac{1}{2},\dfrac{1}{2}\] .
Now, we have to further verify the relationship between the zeros and the coefficients.
As we know, that if the roots of a quadratic polynomial \[a{x^2} + bx + c = 0\] are given, then,
Sum of roots \[ = \] Coefficient of \[x\]/Coefficient of \[{x^2}\] \[ = \dfrac{{ - b}}{a}\]
Product of roots \[ = \] Constant term/Coefficient of \[{x^2}\] \[ = \dfrac{c}{a}\]
Now, the given polynomial is: \[4{s^2} - 4s + 1 = 0\]
Comparing it with \[a{x^2} + bx + c = 0\]
Here, \[a = 4\], \[b = - 4\] and \[c = 1\]
Also, the roots of the polynomial are:
\[s = \dfrac{1}{2}\] and \[s = \dfrac{1}{2}\].
Hence, for verifying part 1 ,
LHS:
Sum of roots \[ = \dfrac{1}{2} + \dfrac{1}{2} = \dfrac{2}{2} = 1\]
RHS:
Coefficient of \[x\]/Coefficient of \[{x^2}\] \[ = \dfrac{{ - b}}{a}\]
Substituting \[a = 4\] and \[b = 4\] in the above expression, we get
Coefficient of \[x\] Coefficient of \[{x^2}\] \[ = \dfrac{{ - \left( { - 4} \right)}}{4} = 1\]
Hence, LHS \[ = \] RHS .
Now, verifying part 2, we get
LHS:
Product of roots \[ = \dfrac{1}{2} \times \dfrac{1}{2} = \dfrac{1}{4}\]
RHS:
Constant term/Coefficient of \[ = \dfrac{c}{a} = \dfrac{1}{4}\]
Hence, LHS \[ = \] RHS
Clearly, both the parts are verified , i.e. LHS = RHS

$\therefore $ We have verified the relationship between the zeros and the coefficients for the given quadratic polynomial \[4{s^2} - 4s + 1 = 0\].

Note:
While finding the roots of the given quadratic polynomial, instead of applying the method of middle term splitting, we could have found its factors by just a little bit of observation
Since, \[4{s^2} - 4s + 1 = 0\]
Now, this can also be written as:
\[ \Rightarrow {\left( {2s} \right)^2} - 2\left( {2s} \right)\left( 1 \right) + {\left( 1 \right)^2} = 0\]
The above is similar to the algebraic identity \[{\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2}\]. So on comparing, we get
\[a = 2s\] and \[b = 1\]
Therefore, simplifying the above equation using identity, we get
\[ \Rightarrow {\left( {2s - 1} \right)^2} = 0\]
Now, this can be written as:
\[\left( {2s - 1} \right)\left( {2s - 1} \right) = 0\]
Hence, \[s = \dfrac{1}{2}\] and $s= \dfrac{1}{2}$.
Therefore, in mathematics, there are various ways to solve a question; the only thing which is required is a keen observation. Else, solving this question was not tough, the only place where we could have gone wrong was the middle term splitting method; hence, it should be solved carefully.