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The number of zeros of the following quadratic polynomial are:

Last updated date: 13th Jun 2024
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Hint: The solutions of an equation are also sometimes called as its zeros or roots.
A polynomial / equation of degree n, has at most n roots.
The two solutions of the general quadratic equation $a{{x}^{2}}+bx+c=0$ , are given as:
 $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
(i)If ${{b}^{2}}-4ac>0$ , then there are two distinct real solutions.
(ii)If ${{b}^{2}}-4ac<0$ , then there are two distinct complex solutions.
(iii)If ${{b}^{2}}-4ac=0$ , then there are two real solutions, each being $x=\dfrac{-b}{2a}$ .

Complete step-by-step answer:
Comparing the given quadratic equation $4{{s}^{2}}-4s+1$ with the general quadratic equation $a{{x}^{2}}+bx+c=0$ , we can say that a = 4, b = -4 and c = 1.
Let us examine the value of ${{b}^{2}}-4ac$ to determine the number of and the nature of the zeros. Using the above values of a, b and c, we get:
 ${{b}^{2}}-4ac={{(-4)}^{2}}-4(4)(1)=16-16=0$ .
Since, ${{b}^{2}}-4ac=0$ , there are two real solutions, which are $s=\dfrac{-b}{2a}=\dfrac{-(-4)}{2(4)}=\dfrac{4}{8}=\dfrac{1}{2}$ .
Even though there are two zeros (solutions), they are both equal.
So, we can say that there is only one zero and its value is $\dfrac{1}{2}$ .

Note: We can also solve the given equation by factoring it using the identity ${{a}^{2}}-2ab+{{b}^{2}}={{(a-b)}^{2}}$ , which will give us only one solution, that is a = b.
The quantity ${{b}^{2}}-4ac$ is also called Discriminant and is denoted by D or Δ.
The degree of a polynomial/equation is the highest power of the variables occurring in it.
(xy) has a degree of two, because the variables x and y are multiplied together.