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.