Question

Show that 4 and 1 are the zeroes of quadratic polynomial p (x) = x2 -5x +4. Also verify the relation between the zeroes and the coefficient of the polynomial.

Answer 1

Answer:

Comparing the given polynomial with ax^2+bx+c,we get

a=1, b=-5, c=4

p(4)=(4)^2-5(4)+4

     =16-20+4

     =0

p(1)=(1)^2-5(1)+4

     =1-5+4

     =0

There,4,1 are the zeroes of polynomial x^2-5x+4

so,we take ∝=4 and β=1

sum of zeroes=∝+β= -b/a

                        =4+1=-(-5/1)

                        =5=5

product of zeroes=∝β=c/a

                               4*1=4/1

                               4=4 (Hence Proved)

Thank u

                                     

Answer 2

${\large{\mathfrak{\underline{\underline{Solution :}}}}}$

$\sf{Given\: polynomial\:-}$

$\sf{p(x) \ = \ x^2 - 5x + 4}$

$\sf{By\:Middle\:Term\: Factorisation}$

$\sf{p(x)\:=\:x^2 - 4x - x + 4}$

$\sf{p(x) \ = \ x(x - 4) - 1 (x - 4)}$

$\sf{p(x) \ = \ (x - 4)(x - 1)}$

$\sf{To \ find \ zeroes, \ p(x) \ should \ be}$
$\sf{equal \ to \ 0.}$

$\sf{So, \ p(x) \ = \ 0}$

$\therefore$

$\sf{(x - 4)(x - 1 ) = 0}$

$\sf{Using \ Zero \ Product \ Rule}$

$\sf{(x - 4) \ = \ 0 \ \ and \ \ (x - 1 ) \ = \ 0}$

$\sf{x \ = \ 4 \ \ and \ \ x = 1}$

$\sf{Zeroes \ are \ 4 \ and \ 1.}$

$\sf{\underline{Verification \ -}}$

$\sf{Let, \ \alpha \ = \ 4 \ \ and \ \ \beta \ = \ 1}$

$\sf{Also,}$

$\sf{From \ the \ given \ Polynomial, \ we \ get}$

$\sf{a \ = \ 1, \ b \ = \ - 5 \ \ and \ \ c \ = \ 4}$

$\sf{Now,}$

$\sf{\underline{Sum \ of \ zeroes, \ S \ \rightarrow}}$

$\sf{S \ = \ \alpha + \beta}$

$\sf{S \ = \ 4 + 1 \ = \ }$${\boxed{\sf{5}}}$

$\sf{Also, \ S \ = \ {\dfrac{-b}{a}}}$

$\sf{So, \ S \ = \ {\dfrac{-(-5)}{1}}}$

$\sf{S \ = \ }$${\boxed{\sf{5}}}$

$\sf{\underline{Product \ of \ zeroes, \ P \ \rightarrow}}$

$\sf{P \ = \ \alpha \beta}$

$\sf{P \ = \ (4)(1) \ = \ }$${\boxed{\sf{4}}}$

$\sf{Also, \ P \ = \ {\dfrac{c}{a}}}$

$\sf{So, \ P \ = \ {\dfrac{4}{1}}}$

$\sf{P \ = \ }$${\boxed{\sf{4}}}$

$\sf{Hence, \ verified}$