Question

Find the zeroes of polynomial P(x)=4x^2-4x+1 and verify the relation between zero and coefficient

Answer 1

SOLUTION :

First of all factorise :

$4 {x}^{2} - 4x + 1$

$\sf = 4x^2 - ( 2 + 2 )x + 1$

$\sf = 4x^2 -2x - 2x + 1$

$\sf = 2x ( 2x - 1 ) - 1 ( 2x - 1 )$

$\sf = ( 2x - 1 ) ( 2x - 1 )$

FOR FINDING ZEROES :

2x - 1 = 0

=> 2x = 1

•°• x = 1/2

Now,

VERIFICATION :

a = 4 ; b = - 4 ; c = 1

! ) Sum of zeroes = - b/a

=> 1/2 + 1/2 = - ( -4 )/4

=> 2/2 = 4/4

=> 1/1 = 1/1

•°• L.H.S = R.H.S

!!) Product of zeroes = c/a

=> 1/2 × 1/2 = 1/4

Hence proved!

$\rule{200}2$

Answer 2

✯✯ QUESTION ✯✯

Find the zeroes of polynomial $P(x)={4x}^{2}-4x+1$ and verify the relation between zero and coefficient..

━━━━━━━━━━━━━━━━━

✰✰ ANSWER ✰✰

$\longrightarrow{P(x)={4x}^{2}-4x+1}$

$\longrightarrow[{4x}^{2}-(2x+2x)+1]$

$\longrightarrow{{4x}^{2}-2x-2x+1}$

$\longrightarrow{2x(2x-1)-1(2x-1)}$

$\longrightarrow{(2x-1) (2x-1)}$

$=\dfrac{1}{2}\:and\:x=\dfrac{1}{2}$ are the zeroes of the polynomial ${4x}^{2}-4x+1$

Now : -

$\longrightarrow{Sum\:of\:Zeroes=\dfrac{Coff.\:of\:x}{Coff.\:of\:{x}^{2}}}$

$\longrightarrow{α+β=\dfrac{-b}{a}}$

$\longrightarrow{\dfrac{1}{2}+\dfrac{1}{2}=-\dfrac{(-4)}{4}}$

$\longrightarrow{\cancel\dfrac{2}{2}=\cancel\dfrac{4}{4}}$

$\longrightarrow{1=1}$

$\large{\boxed{\bold{\bold{\purple{\sf{L.H.S=R.H.S}}}}}}$

Now : -

$\longrightarrow{Product\:of\:Zeroes=\dfrac{Constant\:Term}{Coff.\:of\:{x}^{2}}}$

$\longrightarrow{αβ=\dfrac{c}{a}}$

$\longrightarrow{\dfrac{1}{2}\times\dfrac{1}{2}=\dfrac{1}{4}}$

$\longrightarrow{\dfrac{1}{4}=\dfrac{1}{4}}$

$\large{\boxed{\bold{\bold{\green{\sf{L.H.S=R.H.S}}}}}}$