Question

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

Answer 1

Hint:
According to the question, firstly put the given equation equal to 0. Then factorize using splitting the middle term to calculate the roots. Hence, to verify the relationship calculate the sum of zeroes and product of zeroes.

Complete step by step solution:
Let us assume f(s) be the polynomial.
So, f(s) = \[4{s^2} - 4s + 1\]
The zero of the polynomial means the value of s where \[f(s) = 0\] .
Therefore, \[4{s^2} - 4s + 1 = 0\]
Here, we will find out the roots using splitting the middle term method.
As, this method says that we have to find two numbers whose sum = \[ - 4\] and product = \[1 \times 4 = 4\]
So, we are getting the 2 results
(i) 1 and 4 whose sum is 5 and product is 4.
(ii) \[ - 2\] and \[ - 2\] whose sum is \[ - 4\] and product is 4.
Hence, we will use 2 result to solve this equation that is:
\[4{s^2} - 4s + 1 = 0\]
\[4{s^2} - 2s - 2s + 1 = 0\]
Taking out common from the pairs of 2 we get,
 \[2s\left( {2s - 1} \right) - 1\left( {2s - 1} \right) = 0\]
Taking 2 same factors one time we get,
\[\left( {2s - 1} \right)\left( {2s - 1} \right) = 0\]
By separating out 2 factors,
 \[2s - 1 = 0\]
Here, on simplifying the above equation we get,
So, \[s = \dfrac{1}{2}\]
Similarly, we can calculate second value of s that is \[s = \dfrac{1}{2}\]
Hence, we get \[s = \dfrac{1}{2},\dfrac{1}{2}\]
Therefore, the equation has two roots \[\left( {\alpha ,\beta } \right)\] of the polynomial.
So, \[\alpha = \dfrac{1}{2}\] and \[\beta = \dfrac{1}{2}\]
Comparing the original equation f(s) = \[4{s^2} - 4s + 1\] with \[a{s^2} + bs + c\]
We, compared the values of a, b and c that is \[a = 4,b = - 4,c = 1\]
Now, according to the question we have to verify the relationship between the zeroes and the coefficients is:

1) Sum of zeroes = \[ - \dfrac{\text{coefficient of x}}{\text{coefficient of} {x^2}}\] which means \[\alpha + \beta = - \dfrac{b}{a}\]
On substituting the values in L.H.S we get,
\[\alpha + \beta = \dfrac{1}{2} + \dfrac{1}{2}\]
On taking the L.C.M of the above equation:
\[\alpha + \beta = 1\]
On substituting the values in R.H.S we get,

\[ - \dfrac{b}{a} = - \left( {\dfrac{{ - 4}}{4}} \right)\]
On simplifying the above equation:
 \[ - \dfrac{b}{a} = 1\]
Hence, L.H.S = R.H.S
Therefore, verified the relationship between the sum of zeroes.
2) Product of zeroes = \[\dfrac{\text{constant terms}}{\text{coefficient of} {x^2}}\] which means \[\alpha \times \beta = \dfrac{c}{a}\]
On substituting the values in L.H.S we get,
\[\alpha \times \beta = \dfrac{1}{2} \times \dfrac{1}{2}\]
On multiplying the above equation:
\[\alpha + \beta = \dfrac{1}{4}\]
On substituting the values in R.H.S we get,

\[\dfrac{c}{a} = \dfrac{1}{4}\]
Hence, L.H.S = R.H.S
Therefore, verified the relationship between products of zeroes.

Note:
To solve these types of questions, we must remember that the zeroes are the roots of the polynomial and we equate the equation equal to 0 to find out the roots of the equation. It is important that we should not substitute the value of s equal to 0, and they put \[s = 0\] which commonly mistakes are done by the students.