Question

form the quadratic polynomial whose zeroes are 1 and -1

Answer 1

Answer:

we can form p

a quadratic polynomials with the given two zeros in two methods

1 st method

(x-1)(x+1)

x^2-1 is the polynomial

and the 2nd method is

if a = -1; b= 1

x^2 -( a+b ) + a*b

x^2 -(0) + (-1)

so the polynomial is x^2 - 1

plZz follow me and thank u

Answer 2

❏ SolutioN :

$\setlength{\unitlength}{1.0 cm}}\begin{picture}(12,4)\thicklines\put(1,1){\line(1,0){6.5}}\put(1,1.1){\line(1,0){6.5}}\end{picture}$

$\blacksquare\:\:\footnotesize{\underline{\underline{Given}}}$

$\footnotesize{1st\:zero = 1}$

$\footnotesize{2'nd\:zero = -1}$

$\blacksquare\:\:\footnotesize{\underline{\underline{To\: Find}}}$

$\footnotesize{find \:the\: quadratic\: equation}$

$\setlength{\unitlength}{1.0 cm}}\begin{picture}(12,4)\thicklines\put(1,1){\line(1,0){6.5}}\put(1,1.1){\line(1,0){6.5}}\end{picture}$

$\footnotesize{\text{we know that for a quadratic equation }if\: \alpha\: and\: \beta}$

$\footnotesize{ \text{are the zeroes of the equation then the quadratic equation is :}}$

$\longrightarrow\footnotesize{ x^2-(\alpha+\beta)x+\alpha\beta=0}$

$\footnotesize{\alpha+\beta=(-1)+(1)=0\:and\: \alpha\beta = (-1)(1)=-1 }$

$\therefore\footnotesize{Quadratic \: polynomial\:is \:,}$

$\footnotesize{=x^2-(0)x+(-1)}$

$\footnotesize{\boxed{=x^2-1}}$

$\setlength{\unitlength}{1.0 cm}}\begin{picture}(12,4)\thicklines\put(1,1){\line(1,0){6.5}}\put(1,1.1){\line(1,0){6.5}}\end{picture}$