Question

Find the zeros of the polynomial and verify the relationship between the zeros and coefficients 25x square + 5x

Answer 1

Answer:

Zeroes are 0 and $\frac{-1}{5}$

Step-by-step explanation:

Given: Quadratic Polynomial , 25x² + 5x

To find: Zeroes of Polynomial and Verify the relation between zeroes and coefficient

To find zeroes we equate polynomial with 0

⇒ 25x² + 5x = 0

5x ( 5x + 1 ) = 0

5x = 0  and 5x + 1 = 0

x = 0   and  x = $\frac{-1}{5}$

Therefore, Zeroes are 0 and $\frac{-1}{5}$

let, α = 0 and β = $\frac{-1}{5}$

first relation is sum of zeroes/roots = $\frac{-coefficient\:of\:x}{coefficient\:of\:x^2}$

LHS = α + β = $0+\frac{-1}{5}=\frac{-1}{5}$

RHS = [tex\frac{-b}{a}=\frac{-5}{25}=\frac{-1}{5}[/tex]

LHS = RHS

Hence Verified

Second Relation is Product of zeroes/roots = $\frac{constant\:term}{coefficient\:of\:x^2}$

LHS = αβ = $0\times]frac{-1}{5}=0$

RHS = [tex\frac{c}{a}=\frac{0}{25}=0[/tex]

LHS = RHS

Hence Verified

Answer 2

Answer:

Zeroes are 0 and  

Step-by-step explanation:

Given: Quadratic Polynomial , 25x² + 5x

To find: Zeroes of Polynomial and Verify the relation between zeroes and coefficient

To find zeroes we equate polynomial with 0

⇒ 25x² + 5x = 0

5x ( 5x + 1 ) = 0

5x = 0  and 5x + 1 = 0

x = 0   and  x =  

Therefore, Zeroes are 0 and  

let, α = 0 and β =  

first relation is sum of zeroes/roots =  

LHS = α + β =  

RHS = [tex\frac{-b}{a}=\frac{-5}{25}=\frac{-1}{5}[/tex]

LHS = RHS

Hence Verified

Second Relation is Product of zeroes/roots =  

LHS = αβ =  

RHS = [tex\frac{c}{a}=\frac{0}{25}=0[/tex]

LHS = RHS

Hence Verified