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:First of all split the middle term to factorize the given equation and find its roots. Now verify if the sum of the roots is equal to \[\dfrac{-b}{a}\] and product of the roots is \[\dfrac{c}{a}\] or not where \[a{{x}^{2}}+bc+c\] is our general quadratic equation.

Complete step-by-step answer:
In this question, we have to find the zeroes of the quadratic equation \[4{{s}^{2}}-4s+1\] and verify the relationship between zeroes and coefficients. Before proceeding with the question, let us talk about the zeroes of the quadratic equation.
Zeroes: Zeroes or roots of quadratic equations are the value of variables say x in the quadratic equation \[a{{x}^{2}}+bx+c\] at which the equation becomes zero. If we have \[\alpha \] and \[\beta \] as the roots of the quadratic equation \[a{{x}^{2}}+bx+c\], then we get,
Sum of the roots \[=\alpha +\beta =\dfrac{-b}{a}\]
Product of the roots \[=\alpha \beta =\dfrac{c}{a}\]
Now, let us consider our question. Here we are given a quadratic equation in terms of s, that is,
\[Q\left( s \right)=4{{s}^{2}}-4s+1\]
We can write the middle term of the above equation that is 4s = 2s + 2s. So by substituting the value of 4s, we get,
\[Q\left( s \right)=4{{s}^{2}}-2s-2s+1\]
\[Q\left( s \right)=2s\left( 2s-1 \right)-1\left( 2s-1 \right)\]
By taking out (2s – 1) common from the above equation, we get,
\[Q\left( s \right)=\left( 2s-1 \right)\left( 2s-1 \right)\]
\[Q\left( s \right)={{\left( 2s-1 \right)}^{2}}\]
So, 2s – 1 = 0 and 2s – 1 = 0
\[s=\dfrac{1}{2}\text{ and }s=\dfrac{1}{2}\]
Hence, we get the two roots of the quadratic equation as \[\dfrac{1}{2}\text{ and }\dfrac{1}{2}\] [repeated roots].
We know that for the general quadratic equation of the form \[a{{x}^{2}}+bx+c=0\]
Sum of the roots \[=\dfrac{-b}{a}\]
Product of the roots \[=\dfrac{c}{a}\]
So, by comparing the given equation \[4{{s}^{2}}-4s+1\] with the general quadratic equation \[a{{x}^{2}}+bx+c\], we get, a = 4, b = – 4, and c = 1.
Hence, we get, the sum of the roots \[=\dfrac{-b}{a}=\dfrac{-\left( -4 \right)}{4}=1.....\left( i \right)\]
And product of the roots \[=\dfrac{c}{a}=\dfrac{1}{4}....\left( ii \right)\]
We have found the roots of the given quadratic equation as \[\dfrac{1}{2}\text{ and }\dfrac{1}{2}\]. So, we get,
Sum of the roots \[=\dfrac{1}{2}+\dfrac{1}{2}=1\]
By substituting the sum of the roots in equation (i), we get 1 = 1. And we get,
Product of the roots \[=\dfrac{1}{2}.\dfrac{1}{2}=\dfrac{1}{4}\]
By substituting the product of roots in equation (i), we get,
\[\dfrac{1}{4}=\dfrac{1}{4}\]
Hence, we have verified the relationship between the roots and coefficient of the given quadratic equation.

Note: In this question, students can also find the roots of the quadratic equation by using the quadratic formula that is \[x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\]. Also, if we have \[\alpha \] and \[\beta \] as the roots of the quadratic equation, then we can also write the equation as \[\left( {{x}^{2}}-\left( \alpha +\beta \right)x+\alpha \beta \right)\] in terms of roots of the equation. Also, some students make the mistake while splitting the middle term, so that must be taken care of.