Question

Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients. 4s²-4s+1

Answer 1

Heya friend,
Here is it:

p(s) = 4s² - 4s + 1

Now, we have to do middle term of it,
We get,

4s² - 2s - 2s + 1 
2s (2s -1) -1 (2s -1)
(2s - 1) (2s - 1)

⇒ 2s - 1 = 0    and     2s - 1 = 0
    2s = 1                    2s = 1
    s = 1/2                    s = 1/2

So, the two zeroes are ⇒ 1/2 and 1/2

Let α be 1/2 and β be 1/2

VERIFICATION


So,

Sum of zeroes = α + β
                        = 1/2 + 1/2
                        = 2/2 
                        = 1
and,

Product of zeroes = αβ
                             = 1/2 × 1/2
                             = 1/4  


and, p (s) = 4s² - 4s + 1
where, a = 4, b = - 4 and c = 1
 
α + β = - b/a
         = -(- 4) / 4 
         = 4 / 4
         = 1
and,

 αβ = c / a 
      = 1 / 4 

Hence Verifies !!!


Thanks!
@ Manav

Answer 2

$\large{\underline{\underline{According\:to\:the\:Question}}}$

f(s) = 4s² - 4s - 1

= (2s - 1)²

= (2s - 1)(2s - 1)

(2s - 1) = 0

2s = 1

$\tt{\rightarrow s=\dfrac{1}{2}}$

(2s - 1) = 0

2s = 1

$\tt{\rightarrow s=\dfrac{1}{2}}$

$\tt{\rightarrow\dfrac{1}{2}\;and\;\dfrac{1}{2}}$

★Now

a = 4 , b = -4 and c = 1

★Sum of zeroes = α + β

$\tt{\rightarrow\dfrac{1}{2}+\dfrac{1}{2}}$

= 1

★And,

$\tt{\rightarrow -\dfrac{b}{a}=\dfrac{-(-4)}{4}=1}$

$\tt{\rightarrow\alpha + \beta=-\dfrac{b}{a}}$

★Product of zeroes = α × β

$\tt{\rightarrow\dfrac{1}{2}\times \dfrac{1}{2}=\dfrac{1}{4}}$

$\tt{\rightarrow\dfrac{c}{a}=\dfrac{1}{4}}$

$\tt{\rightarrow\alpha\times \beta=\dfrac{c}{a}}$

$\bold{\boxed{Hence\;verified}}$